Intermediate Algebra

272 soru

Soru 241Soru

The first term of an arithmetic sequence is 12\frac{1}{2}, and the second term is 56\frac{5}{6}. The first term of a geometric sequence is 222^2, and the common ratio is 232^3. Let AA be the third term of the arithmetic sequence, and let GG be the third term of the geometric sequence. If a third value, VV, is defined as 56\frac{5}{6} less than twice GG, what is the value of V+AV + A?

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Cevap: 15373\frac{1537}{3}

Cevap

15373\frac{1537}{3}
The correct answer is 15373\frac{1537}{3} because calculating the third term of the arithmetic sequence yields A=76A = \frac{7}{6} and the third term of the geometric sequence yields G=256G = 256. Translating the relationship for VV yields V=2(256)56=51256V = 2(256) - \frac{5}{6} = 512 - \frac{5}{6}. Finding the sum of VV and AA yields 51256+76=51213=15373512 - \frac{5}{6} + \frac{7}{6} = 512\frac{1}{3} = \frac{1537}{3}.

Adım Adım Çözüm

1
Calculate the common difference dd of the arithmetic sequence.
d=a2a1=5612=13d = a_2 - a_1 = \frac{5}{6} - \frac{1}{2} = \frac{1}{3}
The common difference is the difference between any term and the preceding term in an arithmetic sequence.
2
Calculate the third term AA of the arithmetic sequence.
A=a1+2d=12+2(13)=76A = a_1 + 2d = \frac{1}{2} + 2\left(\frac{1}{3}\right) = \frac{7}{6}
The nn-th term of an arithmetic sequence is given by an=a1+(n1)da_n = a_1 + (n-1)d.
3
Calculate the third term GG of the geometric sequence.
G=g1r2=22(23)2=2226=28=256G = g_1 \cdot r^2 = 2^2 \cdot (2^3)^2 = 2^2 \cdot 2^6 = 2^8 = 256
The nn-th term of a geometric sequence is given by gn=g1rn1g_n = g_1 \cdot r^{n-1}.
4
Set up and solve for VV using the algebraic relationship described.
V=2G56=2(256)56=51256V = 2G - \frac{5}{6} = 2(256) - \frac{5}{6} = 512 - \frac{5}{6}
The phrase '5/6 less than twice G' translates to 2G562G - \frac{5}{6}.
5
Compute the sum of VV and AA.
V+A=(51256)+76=512+26=512+13=15373V + A = \left(512 - \frac{5}{6}\right) + \frac{7}{6} = 512 + \frac{2}{6} = 512 + \frac{1}{3} = \frac{1537}{3}
Substitute the values of VV and AA and simplify the resulting fractional expression.

Anahtar Kavram

Arithmetic and Geometric Sequences and Series
Tahmini Süre:1m 30s
Soru 242Soru

One of the solutions to the quadratic equation 0.5x2+bx6=00.5x^2 + bx - 6 = 0, where bb is a constant, is x=3x = 3. What is the value of the other solution?

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Cevap: -4

Cevap

The other solution to the quadratic equation is -4.
Substituting the given solution x=3x = 3 into the equation yields 0.5(3)2+3b6=00.5(3)^2 + 3b - 6 = 0. Simplifying this expression gives 4.5+3b6=04.5 + 3b - 6 = 0, which leads to 3b=1.53b = 1.5 and thus b=0.5b = 0.5. With b=0.5b = 0.5, the quadratic equation becomes 0.5x2+0.5x6=00.5x^2 + 0.5x - 6 = 0. Multiplying the entire equation by 2 to obtain integer coefficients results in x2+x12=0x^2 + x - 12 = 0. This quadratic factors into (x3)(x+4)=0(x - 3)(x + 4) = 0, which gives the solutions x=3x = 3 and x=4x = -4. Therefore, the other solution is 4-4. Alternatively, using Vieta's formulas, the product of the roots of a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 is equal to c/ac/a. Here, the product of the roots is 6/0.5=12-6 / 0.5 = -12. Since one root is 33, the other root must be 12/3=4-12 / 3 = -4.

Adım Adım Çözüm

1
Substitute the given solution x=3x = 3 into the quadratic equation to find the value of bb.
b=0.5b = 0.5
Since x=3x = 3 is a solution, it must satisfy the equation, allowing us to solve for the unknown coefficient bb.
2
Rewrite the equation using b=0.5b = 0.5 and simplify by multiplying all terms by 2.
x2+x12=0x^2 + x - 12 = 0
Multiplying the equation by 2 eliminates the decimal coefficients, making the quadratic expression easier to factor.
3
Factor the quadratic equation to determine the roots.
x=3x = 3 or x=4x = -4
The equation factors into (x3)(x+4)=0(x - 3)(x + 4) = 0. Solving for xx yields the given root of 3 and the second root of -4.

Anahtar Kavram

Solving quadratic equations by utilizing a known solution to determine unknown coefficients, and applying factoring techniques or root relationships to find the remaining solution.
Tahmini Süre:1m 30s
Soru 243Soru

A geometric sequence consists of positive terms and has a first term of 1212 and a common ratio of rr. An arithmetic sequence has a first term of 55 and a common difference of dd. If the 3rd3\text{rd} term of the geometric sequence is 33 and the 4th4\text{th} term of the arithmetic sequence is 77, what is the value of r+dr + d?

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Cevap: 76\frac{7}{6}

Cevap

76\frac{7}{6}
The correct value is 76\frac{7}{6}. This is found by first calculating the common ratio of the geometric sequence, where 12r2=3r2=14r=1212r^2 = 3 \Rightarrow r^2 = \frac{1}{4} \Rightarrow r = \frac{1}{2}, and the common difference of the arithmetic sequence, where 5+3d=73d=2d=235 + 3d = 7 \Rightarrow 3d = 2 \Rightarrow d = \frac{2}{3}. Adding these two fractions with a common denominator yields 36+46=76\frac{3}{6} + \frac{4}{6} = \frac{7}{6}.

Adım Adım Çözüm

1
Find the common ratio rr of the geometric sequence.
r=12r = \frac{1}{2}
The formula for the nthn\text{th} term of a geometric sequence is gn=g1rn1g_n = g_1 \cdot r^{n-1}. For the 3rd3\text{rd} term, g3=12r2=3g_3 = 12r^2 = 3, which simplifies to r2=14r^2 = \frac{1}{4}. Since the sequence has positive terms, we take the positive square root to get r=12r = \frac{1}{2}.
2
Find the common difference dd of the arithmetic sequence.
d=23d = \frac{2}{3}
The formula for the nthn\text{th} term of an arithmetic sequence is an=a1+(n1)da_n = a_1 + (n-1)d. For the 4th4\text{th} term, a4=5+3d=7a_4 = 5 + 3d = 7, which simplifies to 3d=23d = 2, or d=23d = \frac{2}{3}.
3
Calculate the sum of rr and dd.
r+d=76r + d = \frac{7}{6}
Adding the two values with a common denominator of 66 gives 12+23=36+46=76\frac{1}{2} + \frac{2}{3} = \frac{3}{6} + \frac{4}{6} = \frac{7}{6}.

Anahtar Kavram

Arithmetic and Geometric Sequences and Series
Tahmini Süre:1m 30s
Soru 244Soru

For what value of the constant cc does the quadratic equation x23x+c=0x^2 - 3x + c = 0 have two complex solutions with imaginary parts equal to ±2i\pm 2i?

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Cevap: 6.256.25

Cevap

The value of the constant cc is 6.256.25.
The correct answer is 6.256.25. Applying the quadratic formula to x23x+c=0x^2 - 3x + c = 0 gives solutions of the form 1.5±94c21.5 \pm \frac{\sqrt{9 - 4c}}{2}. Since these solutions are complex with imaginary parts equal to ±2i\pm 2i, the term under the radical must be negative, and the imaginary component is 94c2=2i\frac{\sqrt{9 - 4c}}{2} = 2i. Multiplying both sides by 22 gives 94c=4i\sqrt{9 - 4c} = 4i. Squaring both sides results in 94c=16i29 - 4c = 16i^2. Substituting i2=1i^2 = -1 gives 94c=169 - 4c = -16. Solving for cc yields 4c=25-4c = -25, which simplifies to c=6.25c = 6.25.

Adım Adım Çözüm

1
Apply the quadratic formula to the equation x23x+c=0x^2 - 3x + c = 0.
The solutions are given by x=(3)±(3)24(1)(c)2(1)=1.5±94c2x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(c)}}{2(1)} = 1.5 \pm \frac{\sqrt{9 - 4c}}{2}.
This expresses the solutions in terms of the constant cc so that the imaginary part can be identified.
2
Set the imaginary term of the solutions equal to the given imaginary parts ±2i\pm 2i.
94c2=2i    94c=4i\frac{\sqrt{9 - 4c}}{2} = 2i \implies \sqrt{9 - 4c} = 4i.
The question specifies that the imaginary parts of the two complex solutions are ±2i\pm 2i.
3
Square both sides of the equation to solve for cc.
94c=(4i)2=16i29 - 4c = (4i)^2 = 16i^2. Since i2=1i^2 = -1, this becomes 94c=169 - 4c = -16.
Squaring eliminates the radical and allows for standard algebraic isolation of the variable cc.
4
Solve the linear equation for cc.
4c=25    c=6.25-4c = -25 \implies c = 6.25.
Subtracting 99 from both sides and then dividing by 4-4 isolates the constant cc.

Anahtar Kavram

Solving quadratic equations with complex roots using the quadratic formula and the properties of the imaginary unit.
Soru 245Soru

A sequence of numbers t1,t2,t3,t_1, t_2, t_3, \dots is defined by t1=3t_1 = 3 and tn+1=3tn2nt_{n+1} = 3t_n - 2^n for all integers n1n \geq 1. What is the value of the fourth term, t4t_4?

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Cevap: 43

Cevap

The value of the fourth term is 43.
To find t4t_4, we use the recursive formula tn+1=3tn2nt_{n+1} = 3t_n - 2^n. Substituting n=1n=1 gives t2=3(3)2=7t_2 = 3(3) - 2 = 7. Substituting n=2n=2 gives t3=3(7)4=17t_3 = 3(7) - 4 = 17. Substituting n=3n=3 gives t4=3(17)8=43t_4 = 3(17) - 8 = 43.

Adım Adım Çözüm

1
Calculate the second term, t2t_2, by substituting n=1n = 1 and t1=3t_1 = 3 into the formula tn+1=3tn2nt_{n+1} = 3t_n - 2^n.
t2=3t121=3(3)2=7t_2 = 3t_1 - 2^1 = 3(3) - 2 = 7
To progress to the fourth term, we must first find each preceding term in the sequence.
2
Calculate the third term, t3t_3, by substituting n=2n = 2 and t2=7t_2 = 7 into the formula.
t3=3t222=3(7)4=17t_3 = 3t_2 - 2^2 = 3(7) - 4 = 17
Using the value of the second term allows us to find the third term.
3
Calculate the fourth term, t4t_4, by substituting n=3n = 3 and t3=17t_3 = 17 into the formula.
t4=3t323=3(17)8=43t_4 = 3t_3 - 2^3 = 3(17) - 8 = 43
This completes the recursive process to find the target term.

Anahtar Kavram

Evaluating terms of a sequence defined by a recursive formula.
Tahmini Süre:1m 30s
Soru 246Soru

In the quadratic equation 2x211x+c=02x^2 - 11x + c = 0, where cc is a constant, the ratio of the two real solutions is 3:83:8. What is the value of cc?

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Cevap: 12

Cevap

The value of the constant cc is 1212.
The correct answer is 1212. By representing the roots in the ratio of 3:83:8 as 3r3r and 8r8r, Vieta's formula for the sum of roots (ba-\frac{b}{a}) gives 3r+8r=112    11r=5.5    r=0.53r + 8r = -\frac{-11}{2} \implies 11r = 5.5 \implies r = 0.5. The actual roots are therefore 1.51.5 and 44. Using Vieta's formula for the product of roots (ca\frac{c}{a}) gives (1.5)(4)=c2    6=c2    c=12(1.5)(4) = \frac{c}{2} \implies 6 = \frac{c}{2} \implies c = 12.

Adım Adım Çözüm

1
Represent the roots using the given ratio.
Let the two roots of the quadratic equation be 3r3r and 8r8r.
The ratio of the two solutions is specified as 3:83:8.
2
Apply Vieta's formula for the sum of roots to find the ratio multiplier rr.
3r+8r=112    11r=5.5    r=0.53r + 8r = -\frac{-11}{2} \implies 11r = 5.5 \implies r = 0.5.
For a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0, the sum of the roots is given by ba-\frac{b}{a}.
3
Determine the numerical values of the two roots.
The roots are 3(0.5)=1.53(0.5) = 1.5 and 8(0.5)=48(0.5) = 4.
Substitute the value of r=0.5r = 0.5 back into the expressions for the roots.
4
Apply Vieta's formula for the product of roots to solve for the constant cc.
(1.5)(4)=c2    6=c2    c=12(1.5)(4) = \frac{c}{2} \implies 6 = \frac{c}{2} \implies c = 12.
For a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0, the product of the roots is given by ca\frac{c}{a}.

Anahtar Kavram

Vieta's formulas and the relationship between the roots and coefficients of a quadratic equation

Alternatif Yöntem

Alternatively, you can express the roots using the quadratic formula: x=11±1218c4x = \frac{11 \pm \sqrt{121 - 8c}}{4}. Since the ratio of the smaller root to the larger root is 3:83:8, we set up the equation: 111218c11+1218c=38\frac{11 - \sqrt{121 - 8c}}{11 + \sqrt{121 - 8c}} = \frac{3}{8}. Cross-multiplying gives 8(111218c)=3(11+1218c)    8881218c=33+31218c    55=111218c    5=1218c    25=1218c    8c=96    c=128(11 - \sqrt{121 - 8c}) = 3(11 + \sqrt{121 - 8c}) \implies 88 - 8\sqrt{121 - 8c} = 33 + 3\sqrt{121 - 8c} \implies 55 = 11\sqrt{121 - 8c} \implies 5 = \sqrt{121 - 8c} \implies 25 = 121 - 8c \implies 8c = 96 \implies c = 12.
Tahmini Süre:1m 30s
Soru 247Soru

In the standard (x,y)(x, y) coordinate plane, a circle is defined by the equation x2+y2=25x^2 + y^2 = 25 and a line is defined by the equation 3x+4y=153x + 4y = 15. The line intersects the circle at two points, AA and BB. What is the distance between point AA and point BB?

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Cevap: 8

Cevap

The distance between the two intersection points is 8.
The correct answer is 8. We can find the intersection points by substituting y=153x4y = \frac{15 - 3x}{4} into the circle's equation x2+y2=25x^2 + y^2 = 25, yielding the quadratic equation 5x218x35=05x^2 - 18x - 35 = 0. Solving this gives x=5x = 5 and x=1.4x = -1.4, with corresponding yy-coordinates y=0y = 0 and y=4.8y = 4.8. The distance between (5,0)(5, 0) and (1.4,4.8)(-1.4, 4.8) is (1.45)2+(4.80)2=6.42+4.82=64=8\sqrt{(-1.4 - 5)^2 + (4.8 - 0)^2} = \sqrt{6.4^2 + 4.8^2} = \sqrt{64} = 8. Alternatively, we can use geometry: the distance from the center of the circle (0,0)(0,0) to the line 3x+4y15=03x + 4y - 15 = 0 is d=3(0)+4(0)1532+42=3d = \frac{|3(0) + 4(0) - 15|}{\sqrt{3^2 + 4^2}} = 3. Since the radius of the circle is r=5r = 5, the right triangle formed by the radius, the perpendicular segment, and half the chord has a half-chord length of 5232=4\sqrt{5^2 - 3^2} = 4. Thus, the total chord length is 2×4=82 \times 4 = 8.

Adım Adım Çözüm

1
Express the linear equation in terms of one variable
y=153x4y = \frac{15 - 3x}{4}
This allows for substitution into the equation of the circle.
2
Substitute the expression into the circle's equation and simplify
x2+(153x4)2=2516x2+(22590x+9x2)=40025x290x175=05x218x35=0x^2 + \left(\frac{15 - 3x}{4}\right)^2 = 25 \Rightarrow 16x^2 + (225 - 90x + 9x^2) = 400 \Rightarrow 25x^2 - 90x - 175 = 0 \Rightarrow 5x^2 - 18x - 35 = 0
To create a single quadratic equation in terms of xx representing the intersection points.
3
Solve the quadratic equation for xx
(5x+7)(x5)=0x=5(5x + 7)(x - 5) = 0 \Rightarrow x = 5 or x=1.4x = -1.4
To find the xx-coordinates of the intersection points.
4
Calculate the corresponding yy-coordinates
For x=5x = 5, y=0y = 0, giving point A(5,0)A(5, 0). For x=1.4x = -1.4, y=4.8y = 4.8, giving point B(1.4,4.8)B(-1.4, 4.8).
To determine the exact coordinates of both intersection points.
5
Apply the distance formula to find the length of the segment ABAB
d=(1.45)2+(4.80)2=(6.4)2+4.82=40.96+23.04=64=8d = \sqrt{(-1.4 - 5)^2 + (4.8 - 0)^2} = \sqrt{(-6.4)^2 + 4.8^2} = \sqrt{40.96 + 23.04} = \sqrt{64} = 8
To compute the final distance between the two intersection points.

Anahtar Kavram

Solving systems of linear and quadratic equations to determine intersection points and calculating the distance between coordinates.
Soru 248Soru

A geometric sequence of positive terms has a first term of 99 and a third term of 44. What is the sum of the first 44 terms of this sequence?

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Cevap: 653\frac{65}{3}

Cevap

The correct answer is 653\frac{65}{3}.
To find the sum of the first 4 terms of a geometric sequence with a1=9a_1 = 9 and a3=4a_3 = 4, we first determine the common ratio rr. Since a3=a1r2a_3 = a_1 \cdot r^2, we have 4=9r24 = 9 \cdot r^2, which gives r2=49r^2 = \frac{4}{9}. Because the terms are positive, r=23r = \frac{2}{3}. The first 4 terms are 99, 66, 44, and 83\frac{8}{3}. Summing these terms gives 9+6+4+83=19+83=6539 + 6 + 4 + \frac{8}{3} = 19 + \frac{8}{3} = \frac{65}{3}.

Adım Adım Çözüm

1
Find the common ratio rr of the geometric sequence.
r=23r = \frac{2}{3}
Since the sequence is geometric, the third term is related to the first term by a3=a1r2a_3 = a_1 \cdot r^2. Substituting the given values yields 4=9r24 = 9 \cdot r^2, which simplifies to r2=49r^2 = \frac{4}{9}. Because all terms in the sequence are positive, rr must be positive, so r=49=23r = \sqrt{\frac{4}{9}} = \frac{2}{3}.
2
Calculate the first 4 terms of the sequence.
a1=9a_1 = 9, a2=6a_2 = 6, a3=4a_3 = 4, a4=83a_4 = \frac{8}{3}
Multiply each term by the common ratio r=23r = \frac{2}{3} to find the subsequent term: a1=9a_1 = 9, a2=923=6a_2 = 9 \cdot \frac{2}{3} = 6, a3=623=4a_3 = 6 \cdot \frac{2}{3} = 4, and a4=423=83a_4 = 4 \cdot \frac{2}{3} = \frac{8}{3}.
3
Sum the first 4 terms of the sequence.
Sum = 653\frac{65}{3}
Add the four terms: 9+6+4+83=19+83=573+83=6539 + 6 + 4 + \frac{8}{3} = 19 + \frac{8}{3} = \frac{57}{3} + \frac{8}{3} = \frac{65}{3}.

Anahtar Kavram

Calculating the sum of the first nn terms of a geometric sequence given its first and third terms.
Soru 249Soru

A geometric sequence has a first term of 12\frac{1}{2} and a common ratio of 14\frac{1}{4}. What is the sum of the first 3 terms of this sequence?

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Cevap: 2132\frac{21}{32}

Cevap

The sum of the first 3 terms of this sequence is 2132\frac{21}{32}.
The sum of the first three terms of a geometric sequence is calculated by finding each individual term and then adding them together. The first term is 12\frac{1}{2}. The second term is obtained by multiplying the first term by the common ratio: 12×14=18\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}. The third term is obtained by multiplying the second term by the common ratio: 18×14=132\frac{1}{8} \times \frac{1}{4} = \frac{1}{32}. To add these terms, we find a common denominator of 32: 1632+432+132=2132\frac{16}{32} + \frac{4}{32} + \frac{1}{32} = \frac{21}{32}.

Adım Adım Çözüm

1
Identify the first three terms of the geometric sequence using the formula an=a1rn1a_n = a_1 \cdot r^{n-1}.
The first term a1a_1 is given as 12\frac{1}{2}. The second term is a2=1214=18a_2 = \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}. The third term is a3=1814=132a_3 = \frac{1}{8} \cdot \frac{1}{4} = \frac{1}{32}.
Before calculating the sum, each individual term to be summed must be determined.
2
Find a common denominator to add the three fractional terms.
The least common multiple of the denominators 2, 8, and 32 is 32. Express the terms with this common denominator: a1=1632a_1 = \frac{16}{32}, a2=432a_2 = \frac{4}{32}, and a3=132a_3 = \frac{1}{32}.
Adding fractions requires a common denominator.
3
Sum the adjusted fractions.
1632+432+132=2132\frac{16}{32} + \frac{4}{32} + \frac{1}{32} = \frac{21}{32}.
This yields the total sum of the first three terms.

Anahtar Kavram

Calculating the sum of a finite geometric series by finding and summing individual terms.
Soru 250Soru

A right triangle has a hypotenuse of length x+4x + 4 inches. The lengths of the two legs of the triangle are xx inches and x+2x + 2 inches. What is the value of xx?

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Cevap: 6

Cevap

6
The correct answer is 6 because applying the Pythagorean theorem yields the relation x2+(x+2)2=(x+4)2x^2 + (x + 2)^2 = (x + 4)^2. Expanding the binomials gives x2+x2+4x+4=x2+8x+16x^2 + x^2 + 4x + 4 = x^2 + 8x + 16, which simplifies to the quadratic equation x24x12=0x^2 - 4x - 12 = 0. Factoring this equation yields (x6)(x+2)=0(x - 6)(x + 2) = 0. Discarding the negative solution x=2x = -2 because length must be positive leaves the correct solution of 6.

Adım Adım Çözüm

1
Set up the equation using the Pythagorean theorem, where the sum of the squares of the legs equals the square of the hypotenuse.
x2+(x+2)2=(x+4)2x^2 + (x + 2)^2 = (x + 4)^2
The sides of a right triangle must satisfy the Pythagorean relation a2+b2=c2a^2 + b^2 = c^2.
2
Expand the squared binomial terms on both sides of the equation.
x2+(x2+4x+4)=x2+8x+16x^2 + (x^2 + 4x + 4) = x^2 + 8x + 16
Applying the algebraic identity (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2 is necessary to simplify the terms.
3
Simplify the equation and move all terms to one side to set the quadratic expression to zero.
x24x12=0x^2 - 4x - 12 = 0
Standard form (ax2+bx+c=0ax^2 + bx + c = 0) is required to solve quadratic equations.
4
Factor the quadratic equation.
(x6)(x+2)=0(x - 6)(x + 2) = 0
Factoring allows finding the roots by setting each linear binomial factor to zero.
5
Solve for xx and discard any physically impossible negative values.
x=6x = 6 (since x=2x = -2 is discarded)
A physical measurement of side length must be strictly positive.

Anahtar Kavram

Formulating and solving quadratic equations derived from the Pythagorean theorem by expanding binomials and factoring.
Soru 251Soru

A circle in the standard (x,y)(x, y) coordinate plane is described by the equation x2+y2=68x^2 + y^2 = 68. A line is described by the equation y=x6y = x - 6. The line intersects the circle at two points. What is the distance between these two points of intersection?

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Cevap: 10210\sqrt{2}

Cevap

The distance between the two points of intersection is 10210\sqrt{2}.
To find the points of intersection, substitute y=x6y = x - 6 into x2+y2=68x^2 + y^2 = 68, which yields x2+(x6)2=68x^2 + (x - 6)^2 = 68. Expanding the binomial correctly results in x2+x212x+36=68x^2 + x^2 - 12x + 36 = 68. Combining like terms and writing the equation in standard form gives 2x212x32=02x^2 - 12x - 32 = 0. Dividing the entire equation by 2 gives x26x16=0x^2 - 6x - 16 = 0. Factoring this quadratic equation yields (x8)(x+2)=0(x - 8)(x + 2) = 0, giving solutions x=8x = 8 and x=2x = -2. Substituting these values back into the linear equation gives the points (8,2)(8, 2) and (2,8)(-2, -8). Finally, using the distance formula, the distance between the two points is (8(2))2+(2(8))2=102+102=200=102\sqrt{(8 - (-2))^2 + (2 - (-8))^2} = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2}.

Adım Adım Çözüm

1
Substitute the linear equation into the circle equation.
x2+(x6)2=68x^2 + (x - 6)^2 = 68
To eliminate yy and solve for the xx-coordinates of the intersection points.
2
Expand the binomial and collect like terms.
2x212x32=02x^2 - 12x - 32 = 0
Expanding (x6)2(x - 6)^2 yields x212x+36x^2 - 12x + 36, and combining it with x2x^2 and subtracting 6868 from both sides puts the equation in quadratic form.
3
Divide the quadratic equation by 2 and solve by factoring.
(x8)(x+2)=0(x - 8)(x + 2) = 0, so x=8x = 8 or x=2x = -2
Simplifying the equation to x26x16=0x^2 - 6x - 16 = 0 makes it easy to find the roots of the quadratic equation.
4
Find the corresponding yy-coordinates by substituting the xx-values back into the linear equation.
For x=8x = 8, y=2y = 2, yielding point (8,2)(8, 2). For x=2x = -2, y=8y = -8, yielding point (2,8)(-2, -8).
To determine the full coordinates of the two intersection points.
5
Apply the distance formula to find the distance between (8,2)(8, 2) and (2,8)(-2, -8).
d=(8(2))2+(2(8))2=102+102=200=102d = \sqrt{(8 - (-2))^2 + (2 - (-8))^2} = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2}
To calculate the straight-line distance between the two coordinate points.

Anahtar Kavram

Solving systems of linear and non-linear equations by substitution and finding the distance between intersection points.
Tahmini Süre:1m 30s
Soru 252Soru

A toy rocket is launched upward from a platform. Its height hh, in meters, above the ground tt seconds after launch is modeled by the function h(t)=4.9t2+7.35t+12.25h(t) = -4.9t^2 + 7.35t + 12.25. According to this model, how many seconds after launch does the rocket strike the ground?

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Cevap: 2.5

Cevap

The rocket strikes the ground 2.52.5 seconds after launch.
The correct answer of 2.52.5 is determined by setting the height h(t)h(t) to 00 and solving the resulting quadratic equation using the quadratic formula. Since time must be non-negative in this physical context, the negative solution of 1-1 is discarded, leaving 2.52.5 seconds as the time when the rocket strikes the ground.

Adım Adım Çözüm

1
Set the height function h(t)h(t) to 00.
4.9t2+7.35t+12.25=0-4.9t^2 + 7.35t + 12.25 = 0
The rocket strikes the ground when its height above the ground is 00 meters.
2
Apply the quadratic formula t=b±b24ac2at = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
t=7.35±(7.35)24(4.9)(12.25)2(4.9)t = \frac{-7.35 \pm \sqrt{(7.35)^2 - 4(-4.9)(12.25)}}{2(-4.9)}
This formula provides the solutions to any quadratic equation of the form at2+bt+c=0at^2 + bt + c = 0.
3
Calculate the discriminant and its square root.
b24ac=294.1225b^2 - 4ac = 294.1225 and 294.1225=17.15\sqrt{294.1225} = 17.15
Evaluating the term under the radical simplifies the quadratic formula expression.
4
Evaluate the two possible values for tt.
t=1t = -1 or t=2.5t = 2.5
Solving the simplified expression gives the two mathematical roots of the quadratic equation.
5
Choose the physically valid solution.
t=2.5t = 2.5
Time must be positive in this scenario, so the negative solution t=1t = -1 is discarded.

Anahtar Kavram

Solving a quadratic equation with decimal coefficients using the quadratic formula in a real-world motion context.
Soru 253Soru

A quadratic equation is defined by x2bx+18=0x^2 - bx + 18 = 0, where bb is a positive constant. If the difference between the two real solutions of this equation is 3, what is the value of bb?

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Cevap: 9

Cevap

The value of the positive constant bb is 9.
By applying the quadratic formula, the roots of the equation are x=b±b2722x = \frac{b \pm \sqrt{b^2 - 72}}{2}. The difference between these roots is b272\sqrt{b^2 - 72}. Setting this equal to 3 gives b272=3\sqrt{b^2 - 72} = 3. Squaring both sides yields b272=9b^2 - 72 = 9, which simplifies to b2=81b^2 = 81. Taking the positive root since bb is a positive constant gives b=9b = 9.

Adım Adım Çözüm

1
Express the roots of the quadratic equation x2bx+18=0x^2 - bx + 18 = 0 using the quadratic formula.
The roots are x=b±b24(1)(18)2=b±b2722x = \frac{b \pm \sqrt{b^2 - 4(1)(18)}}{2} = \frac{b \pm \sqrt{b^2 - 72}}{2}.
This provides a formulaic representation of the two solutions in terms of the unknown parameter bb.
2
Subtract the smaller root from the larger root to represent the difference between the solutions, and set this expression equal to 3.
Difference =b+b2722bb2722=b272=3= \frac{b + \sqrt{b^2 - 72}}{2} - \frac{b - \sqrt{b^2 - 72}}{2} = \sqrt{b^2 - 72} = 3.
The problem specifies that the difference between the two real solutions is 3.
3
Square both sides of the equation to eliminate the radical, and solve for the positive constant bb.
b272=9b2=81b=9b^2 - 72 = 9 \Rightarrow b^2 = 81 \Rightarrow b = 9 (since bb is positive).
Squaring both sides allows us to isolate b2b^2 and find the value of bb that satisfies the initial condition.

Anahtar Kavram

Solving for quadratic coefficients using the difference of roots derived from the quadratic formula.

Alternatif Yöntem

Use Vieta's formulas. Let the roots be r1r_1 and r2r_2. We know that r1+r2=br_1 + r_2 = b and r1r2=18r_1 r_2 = 18. We are given that the difference between the roots is 3, so r1r2=3|r_1 - r_2| = 3. We can use the algebraic identity (r1r2)2=(r1+r2)24r1r2(r_1 - r_2)^2 = (r_1 + r_2)^2 - 4r_1 r_2. Substituting the known values gives 32=b24(18)3^2 = b^2 - 4(18), which simplifies to 9=b2729 = b^2 - 72, leading to b2=81b^2 = 81. Since b>0b > 0, we find b=9b = 9.
Tahmini Süre:1m 30s
Soru 254Soru

In the standard (x,y)(x, y) coordinate system, a line given by the equation y=2x+4y = 2x + 4 intersects a parabola given by the equation y=x22x1y = x^2 - 2x - 1 at exactly two points. What is the sum of the yy-coordinates of these two points of intersection?

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Cevap: 16

Cevap

16
The system of equations is solved by setting the equations equal to each other, yielding the quadratic equation x24x5=0x^2 - 4x - 5 = 0. Solving for xx gives x=5x = 5 and x=1x = -1. Substituting these values into the linear equation gives the y-coordinates 1414 and 22. The sum of these y-coordinates is 14+2=1614 + 2 = 16.

Adım Adım Çözüm

1
Equate the linear and quadratic equations to find the x-values of the intersection points.
x22x1=2x+4x^2 - 2x - 1 = 2x + 4
At the points of intersection, the y-values of both equations must be equal.
2
Rearrange the equation to standard quadratic form.
x24x5=0x^2 - 4x - 5 = 0
Grouping all terms on one side allows the quadratic equation to be solved.
3
Factor the quadratic equation to find the x-coordinates.
(x5)(x+1)=0(x - 5)(x + 1) = 0, so x=5x = 5 or x=1x = -1
The numbers that multiply to 5-5 and add up to 4-4 are 5-5 and 11.
4
Substitute the x-coordinates into the linear equation to find the y-coordinates.
For x=5x = 5: y=2(5)+4=14y = 2(5) + 4 = 14. For x=1x = -1: y=2(1)+4=2y = 2(-1) + 4 = 2.
Evaluating the linear equation is simpler than evaluating the quadratic equation.
5
Find the sum of the y-coordinates.
14+2=1614 + 2 = 16
The question asks for the sum of the y-coordinates of the two points of intersection.

Anahtar Kavram

Solving a system of linear and quadratic equations by substitution.

Alternatif Yöntem

Instead of solving for the individual intersection points, Vieta's formulas can be applied. The x-coordinates satisfy x24x5=0x^2 - 4x - 5 = 0, so their sum is x1+x2=4x_1 + x_2 = 4. Since the points lie on the line y=2x+4y = 2x + 4, the sum of the y-coordinates is y1+y2=(2x1+4)+(2x2+4)=2(x1+x2)+8=2(4)+8=16y_1 + y_2 = (2x_1 + 4) + (2x_2 + 4) = 2(x_1 + x_2) + 8 = 2(4) + 8 = 16.
Tahmini Süre:1m 30s
Soru 255Soru

A model glider is launched from a hill. Its height h(t)h(t), in meters above the valley floor tt seconds after launch, is modeled by the function h(t)=0.5t2+3.5t+10h(t) = -0.5t^2 + 3.5t + 10. Which of the following is a possible value of tt, in seconds, when the glider is at a height of exactly 1212 meters?

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Cevap: 7+332\frac{7 + \sqrt{33}}{2}

Cevap

7+332\frac{7 + \sqrt{33}}{2}
The correct answer is obtained by setting the height equation h(t)=12h(t) = 12, which simplifies to 0.5t2+3.5t2=0-0.5t^2 + 3.5t - 2 = 0. Multiplying by 2-2 gives the standard form t27t+4=0t^2 - 7t + 4 = 0. Applying the quadratic formula with a=1a = 1, b=7b = -7, and c=4c = 4 gives t=7±332t = \frac{7 \pm \sqrt{33}}{2}. Thus, the option representing 7+332\frac{7 + \sqrt{33}}{2} is the correct choice.

Adım Adım Çözüm

1
Set the height function equal to the target height of 12 meters.
0.5t2+3.5t+10=12-0.5t^2 + 3.5t + 10 = 12
To find when the glider reaches exactly 12 meters, we set the model function equal to 12.
2
Subtract 12 from both sides to set the quadratic equation to zero.
0.5t2+3.5t2=0-0.5t^2 + 3.5t - 2 = 0
A quadratic equation must be in standard form at2+bt+c=0at^2 + bt + c = 0 before applying the quadratic formula.
3
Multiply the entire equation by 2-2 to eliminate decimal coefficients.
t27t+4=0t^2 - 7t + 4 = 0
Working with integer coefficients reduces calculation errors when applying the quadratic formula. Here, a=1a = 1, b=7b = -7, and c=4c = 4.
4
Apply the quadratic formula t=b±b24ac2at = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
t=(7)±(7)24(1)(4)2(1)=7±49162=7±332t = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(4)}}{2(1)} = \frac{7 \pm \sqrt{49 - 16}}{2} = \frac{7 \pm \sqrt{33}}{2}
The quadratic formula is used to solve quadratic equations that cannot be easily factored using integers.

Anahtar Kavram

Solving quadratic equations with decimal coefficients by converting to standard integer form and applying the quadratic formula.
Soru 256Soru

For what value of cc does the quadratic equation 0.5x23x+c=00.5x^2 - 3x + c = 0 have two real solutions that differ by exactly 4?

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Cevap: 2.5

Cevap

2.5
The correct value is 2.5. By utilizing the formula for the difference of the roots, b24aca=4\frac{\sqrt{b^2 - 4ac}}{|a|} = 4, and substituting a=0.5a = 0.5 and b=3b = -3, we get 92c0.5=4\frac{\sqrt{9 - 2c}}{0.5} = 4. This simplifies to 92c=2\sqrt{9 - 2c} = 2, which squares to 92c=49 - 2c = 4. Solving for cc yields 2.5.

Adım Adım Çözüm

1
Identify the coefficients and apply the relationship for the difference between two roots.
For a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0, the roots x1x_1 and x2x_2 satisfy x1x2=b24aca|x_1 - x_2| = \frac{\sqrt{b^2 - 4ac}}{|a|}. Here, a=0.5a = 0.5, b=3b = -3, and the difference is 4.
This formula relates the difference of the roots directly to the coefficients of the quadratic equation.
2
Substitute the given values into the formula and solve for cc.
Substituting the values gives (3)24(0.5)c0.5=492c0.5=4\frac{\sqrt{(-3)^2 - 4(0.5)c}}{|0.5|} = 4 \Rightarrow \frac{\sqrt{9 - 2c}}{0.5} = 4. Multiplying both sides by 0.5 yields 92c=2\sqrt{9 - 2c} = 2. Squaring both sides gives 92c=49 - 2c = 4.
Simplifying the equation isolates the variable cc under the radical.
3
Complete the algebraic isolation to find the final value of cc.
2c=5c=2.52c = 5 \Rightarrow c = 2.5.
This final step solves the linear equation for cc.

Anahtar Kavram

Using the discriminant and properties of roots to solve quadratic equations with given constraints.
Soru 257Soru

The path of a particle in the standard (x,y)(x, y) coordinate plane is described by the linear equation 3xy=23x - y = 2, and the path of another particle is described by the quadratic equation y=x2x7y = x^2 - x - 7. If the two paths intersect at two locations, what is the sum of the yy-coordinates of these intersection points?

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Cevap: 8

Cevap

The sum of the yy-coordinates of the intersection points is 8.
The correct answer is 8. Solving the system by setting x2x7=3x2x^2 - x - 7 = 3x - 2 results in the quadratic equation x24x5=0x^2 - 4x - 5 = 0. Factoring gives (x5)(x+1)=0(x-5)(x+1) = 0, which yields intersection xx-coordinates of x=5x = 5 and x=1x = -1. Substituting these back into the linear equation gives yy-coordinates of 1313 and 5-5. Adding these values together yields 13+(5)=813 + (-5) = 8.

Adım Adım Çözüm

1
Express the linear equation in terms of yy.
y=3x2y = 3x - 2
This allows for direct substitution into the quadratic equation.
2
Equate the linear and quadratic expressions to solve for the xx-coordinates of the intersection points.
x2x7=3x2x^2 - x - 7 = 3x - 2
Intersection points share the same coordinates for both equations.
3
Set the quadratic equation to zero.
x24x5=0x^2 - 4x - 5 = 0
This puts the equation in standard form so it can be solved by factoring.
4
Factor the quadratic equation.
(x5)(x+1)=0(x - 5)(x + 1) = 0, so x=5x = 5 or x=1x = -1
Factoring determines the xx-coordinates of the intersection points.
5
Substitute the xx-values into the linear equation to determine the yy-coordinates.
For x=5x = 5, y=13y = 13. For x=1x = -1, y=5y = -5.
Finding the yy-coordinates is necessary to compute their sum.
6
Add the yy-coordinates together.
13+(5)=813 + (-5) = 8
The question asks for the sum of the yy-coordinates of the intersection points.

Anahtar Kavram

Solving systems of linear and quadratic equations by substitution and factoring

Alternatif Yöntem

We can use Vieta's formulas to find the sum of the yy-coordinates without calculating each individual coordinate. The sum of the yy-coordinates is y1+y2=(3x12)+(3x22)=3(x1+x2)4y_1 + y_2 = (3x_1 - 2) + (3x_2 - 2) = 3(x_1 + x_2) - 4. Since x1x_1 and x2x_2 are the roots of x24x5=0x^2 - 4x - 5 = 0, Vieta's formulas state that the sum of the roots is x1+x2=41=4x_1 + x_2 = -\frac{-4}{1} = 4. Substituting this value into our sum expression yields 3(4)4=124=83(4) - 4 = 12 - 4 = 8.
Tahmini Süre:1m 30s
Soru 258Soru

A parabolic arch is modeled by the equation y=(x3)25y = (x - 3)^2 - 5 in the standard (x,y)(x, y) coordinate plane. A straight pathway is modeled by a line where the yy-coordinate of any point is 2 less than its xx-coordinate. If the pathway intersects the arch at points AA and BB, what is the area, in square units, of the triangle with vertices at AA, BB, and the origin (0,0)(0, 0)?

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Cevap: 5

Cevap

The area of the triangle with vertices at the intersection points and the origin is 5 square units.
To find the area of the triangle, we first solve the system of equations. Substituting the pathway's equation y = x - 2 into the parabola's equation y = (x - 3)^2 - 5 yields x^2 - 7x + 6 = 0, which factors to (x - 6)(x - 1) = 0. This gives x = 6 and x = 1. The corresponding y-coordinates are y = 4 and y = -1, representing the intersection points (6, 4) and (1, -1). The area of the triangle with these vertices and the origin (0, 0) is calculated as 0.5 * |6(-1) - 4(1)| = 5.

Adım Adım Çözüm

1
Set up the system of equations by substituting the linear equation into the quadratic equation.
(x3)25=x2(x - 3)^2 - 5 = x - 2
The pathway is described as having a y-coordinate that is 2 less than the x-coordinate, which translates to the linear equation y = x - 2. Substituting this into the parabola's equation allows us to find the intersection points.
2
Expand the quadratic term and simplify the equation into standard quadratic form.
x27x+6=0x^2 - 7x + 6 = 0
Expanding (x3)2(x - 3)^2 gives x26x+9x^2 - 6x + 9. Subtracting xx and adding 22 to both sides results in standard quadratic form.
3
Factor the quadratic equation to solve for the x-coordinates.
(x6)(x1)=0(x - 6)(x - 1) = 0, so x=6x = 6 or x=1x = 1
Factoring allows us to find the x-values that satisfy the intersection condition.
4
Substitute the x-values back into the linear equation to find the corresponding y-coordinates.
For x=6x = 6, y=4y = 4 giving point (6,4)(6, 4). For x=1x = 1, y=1y = -1 giving point (1,1)(1, -1).
The intersection points must satisfy both equations in the system.
5
Calculate the area of the triangle with vertices (0,0)(0, 0), (6,4)(6, 4), and (1,1)(1, -1) using the coordinate area formula.
Area = 126(1)4(1)=1210=5\frac{1}{2} |6(-1) - 4(1)| = \frac{1}{2} |-10| = 5 square units.
The area of a triangle with one vertex at the origin and others at (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is 12x1y2y1x2\frac{1}{2} |x_1 y_2 - y_1 x_2|.

Anahtar Kavram

Solving systems of linear and quadratic equations and finding the area of a triangle in the coordinate plane.
Soru 259Soru

For the quadratic equation 0.25x21.5x+c=00.25x^2 - 1.5x + c = 0, the discriminant is equal to 44. What is the value of the constant cc?

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Cevap: 1.75-1.75

Cevap

The value of the constant cc is 1.75-1.75.
The correct option is 1.75-1.75. To find the value of cc, substitute the coefficients a=0.25a = 0.25, b=1.5b = -1.5, and the discriminant D=4D = 4 into the formula D=b24acD = b^2 - 4ac. This gives 4=(1.5)24(0.25)c4 = (-1.5)^2 - 4(0.25)c, which simplifies to 4=2.25c4 = 2.25 - c. Solving for cc yields c=2.254=1.75c = 2.25 - 4 = -1.75.

Adım Adım Çözüm

1
Identify the values of the coefficients from the quadratic equation 0.25x21.5x+c=00.25x^2 - 1.5x + c = 0.
The coefficients are a=0.25a = 0.25, b=1.5b = -1.5, and the constant is cc.
These coefficients are required to compute the discriminant.
2
Recall the formula for the discriminant DD and substitute the known values, including the given discriminant D=4D = 4.
4=(1.5)24(0.25)c4 = (-1.5)^2 - 4(0.25)c
This sets up an equation to solve for the unknown constant cc.
3
Simplify the squared term and the multiplication of the coefficients.
4=2.251c4 = 2.25 - 1c, which simplifies to 4=2.25c4 = 2.25 - c.
Squaring 1.5-1.5 yields 2.252.25, and 4(0.25)=14(0.25) = 1.
4
Solve the linear equation for cc.
c=1.75c = -1.75
Subtracting 2.252.25 from both sides gives 1.75=c1.75 = -c, which means c=1.75c = -1.75.

Anahtar Kavram

Using the discriminant formula D=b24acD = b^2 - 4ac to solve for an unknown coefficient in a quadratic equation.
Tahmini Süre:1m 30s
Soru 260Soru

The quadratic equation 1.5x2kx+6=01.5x^2 - kx + 6 = 0, where kk is a positive constant, has exactly one real solution. What is the value of kk?

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Cevap: 6

Cevap

6
For the quadratic equation 1.5x2kx+6=01.5x^2 - kx + 6 = 0 to have exactly one real solution, the discriminant b24acb^2 - 4ac must equal 00. Substituting a=1.5a = 1.5, b=kb = -k, and c=6c = 6 gives (k)24(1.5)(6)=k236=0(-k)^2 - 4(1.5)(6) = k^2 - 36 = 0, which yields k2=36k^2 = 36. Since kk must be a positive constant, kk must be 66.

Adım Adım Çözüm

1
Identify the coefficients of the quadratic equation 1.5x2kx+6=01.5x^2 - kx + 6 = 0.
a=1.5a = 1.5, b=kb = -k, and c=6c = 6
To use the discriminant formula, we need to know the values of aa, bb, and cc from the standard form ax2+bx+c=0ax^2 + bx + c = 0.
2
Set the discriminant equal to zero.
b24ac=0b^2 - 4ac = 0
A quadratic equation has exactly one real solution if and only if its discriminant is equal to zero.
3
Substitute the coefficients into the discriminant formula and simplify.
k236=0k^2 - 36 = 0
Substituting a=1.5a = 1.5, b=kb = -k, and c=6c = 6 into the formula gives (k)24(1.5)(6)=k236=0(-k)^2 - 4(1.5)(6) = k^2 - 36 = 0.
4
Solve the equation for the positive constant kk.
k=6k = 6
Solving k2=36k^2 = 36 gives k=6k = 6 or k=6k = -6. Since the problem states that kk is a positive constant, we choose k=6k = 6.

Anahtar Kavram

Determining the number of real solutions of a quadratic equation using the discriminant
ÖncekiSayfa 13 / 14Sonraki
Intermediate Algebra Alıştırma Soruları — ACT — Sayfa 13 | Examkin