Tüm alıştırma soruları

541 soru

Soru 221Soru

In the standard (x,y)(x,y) coordinate plane, a region in the first quadrant is bounded by the xx-axis, the yy-axis, and the line with equation ax+by=cax + by = c, where aa, bb, and cc are positive constants. The line passes through the point (8,18)(8, 18). If the area of this region is minimized when a=3a = 3, what is the value of cc?

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Cevap: 48

Cevap

48
Substituting the given point and a=3a = 3 into the equation yields c=24+18bc = 24 + 18b. The area of the triangle formed by the intercepts is A=c26bA = \frac{c^2}{6b}. Substituting cc gives A=6(9b+24+16b)A = 6(9b + 24 + \frac{16}{b}). Using AM-GM, the minimum occurs when 9b=16b9b = \frac{16}{b}, resulting in b=43b = \frac{4}{3}. Using this value, we find c=48c = 48.

Adım Adım Çözüm

1
Substitute the point (8,18)(8, 18) and a=3a = 3 into the equation ax+by=cax + by = c.
24+18b=c24 + 18b = c
This establishes a relationship between the constants bb and cc based on the given point that lies on the line.
2
Calculate the xx-intercept and yy-intercept of the line.
xx-intercept is at x=c3x = \frac{c}{3}, and yy-intercept is at y=cby = \frac{c}{b}.
The boundary of the region in the first quadrant is defined by these coordinate intercepts.
3
Formulate the area AA of the right triangle bounded by the axes and the line.
A=c26bA = \frac{c^2}{6b}
The area of a right triangle with vertices at the origin and the intercepts is 12baseheight\frac{1}{2} \cdot \text{base} \cdot \text{height}.
4
Substitute c=24+18bc = 24 + 18b into the area formula and simplify.
A=6(9b+24+16b)A = 6\left(9b + 24 + \frac{16}{b}\right)
Expressing the area as a single-variable function of bb allows us to find its minimum value.
5
Apply the AM-GM inequality to minimize the variable term 9b+16b9b + \frac{16}{b}.
b=43b = \frac{4}{3} minimizes the expression.
The sum of two positive terms is minimized when the terms are equal, so 9b=16b    b2=169    b=439b = \frac{16}{b} \implies b^2 = \frac{16}{9} \implies b = \frac{4}{3}.
6
Calculate the value of cc using the minimizing value of bb.
c=48c = 48
Substituting b=43b = \frac{4}{3} into the relation c=24+18bc = 24 + 18b yields the constant value cc for the minimum area.

Anahtar Kavram

Minimizing the area bounded by a line and the coordinate axes using linear equation forms and algebraic minimization.

Alternatif Yöntem

Instead of using the AM-GM inequality, you can find the minimum by taking the derivative of the area function A(b)=54b+144+96bA(b) = 54b + 144 + \frac{96}{b} with respect to bb. Setting the derivative A(b)=5496b2=0A'(b) = 54 - \frac{96}{b^2} = 0 yields b2=9654=169b^2 = \frac{96}{54} = \frac{16}{9}, which gives b=43b = \frac{4}{3} for b>0b > 0.
Tahmini Süre:3m 0s
Soru 222Soru

In the standard (x,y)(x, y) coordinate plane, an ellipse is centered at the origin (0,0)(0, 0) and has vertices at (5,0)(-5, 0) and (5,0)(5, 0). If the distance between the two foci of the ellipse is 88, what is the length of the minor axis of the ellipse?

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Cevap: 6

Cevap

The correct answer is 6.
The correct answer is 6 because the ellipse has a horizontal major axis with a=5a = 5 and focal distance c=4c = 4. Using the relationship c2=a2b2c^2 = a^2 - b^2, we solve for the semi-minor axis bb to get b=3b = 3. The total length of the minor axis is 2b=62b = 6.

Adım Adım Çözüm

1
Determine the semi-major axis length aa.
a=5a = 5
The vertices are at (±5,0)(\pm 5, 0), which are 55 units from the center (0,0)(0, 0) along the major axis.
2
Determine the distance from the center to each focus cc.
c=4c = 4
The distance between the two foci is 2c=82c = 8, so the distance from the center to a focus is c=4c = 4.
3
Find the semi-minor axis length bb.
b=3b = 3
Using the relation c2=a2b2c^2 = a^2 - b^2 for ellipses, we get 42=52b2    b2=9    b=34^2 = 5^2 - b^2 \implies b^2 = 9 \implies b = 3.
4
Calculate the full length of the minor axis.
66
The length of the minor axis is 2b=2(3)=62b = 2(3) = 6.

Anahtar Kavram

The relationship between the semi-major axis, semi-minor axis, and focal distance of an ellipse.
Tahmini Süre:1m 30s
Soru 223Soru

Given the functions f(x)=3x+4f(x) = \sqrt{3x + 4} and g(x)=x25g(x) = x^2 - 5, what is the value of g(f(7))g(f(7))?

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Cevap: 20

Cevap

20
Evaluating the inner function first gives f(7)=3(7)+4=5f(7) = \sqrt{3(7) + 4} = 5. Substituting this value into the outer function yields g(5)=525=20g(5) = 5^2 - 5 = 20.

Adım Adım Çözüm

1
Evaluate the inner function f(x)f(x) at x=7x = 7.
f(7)=5f(7) = 5
To evaluate a composite function of the form g(f(x))g(f(x)), we first calculate the value of the inner function f(x)f(x) at the given input.
2
Substitute the output of the inner function as the input for the outer function g(x)g(x).
g(5)=20g(5) = 20
Since f(7)=5f(7) = 5, evaluating g(f(7))g(f(7)) is equivalent to evaluating g(5)g(5).

Anahtar Kavram

Function composition involves evaluating an inner function and then using that result as the input for an outer function.
Soru 224Soru

If xx and yy are non-zero real numbers such that the expression (x2ya)3x5y2\frac{(x^2 y^a)^3}{x^5 y^{-2}} is equivalent to xby14x^b y^{14} for some integers aa and bb, what is the value of a+ba + b?

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Cevap: 5

Cevap

The value of a+ba + b is 55.
Simplifying the expression (x2ya)3x5y2\frac{(x^2 y^a)^3}{x^5 y^{-2}} using the exponent rules yields x65y3a(2)=x1y3a+2x^{6-5} y^{3a-(-2)} = x^1 y^{3a+2}. Equating this to xby14x^b y^{14} shows that b=1b = 1 and 3a+2=143a + 2 = 14. Solving for aa gives a=4a = 4. Thus, the sum a+ba + b is 4+1=54 + 1 = 5.

Adım Adım Çözüm

1
Simplify the numerator of the given expression.
x6y3ax^6 y^{3a}
Applying the power of a product rule (uv)n=unvn(uv)^n = u^n v^n and the power of a power rule (um)n=umn(u^m)^n = u^{mn} to (x2ya)3(x^2 y^a)^3 results in (x2)3(ya)3=x6y3a(x^2)^3 (y^a)^3 = x^6 y^{3a}.
2
Simplify the quotient by subtracting exponents with the same base.
x1y3a+2x^1 y^{3a+2}
Using the quotient rule umun=umn\frac{u^m}{u^n} = u^{m-n}, the base xx term becomes x65=x1x^{6-5} = x^1, and the base yy term becomes y3a(2)=y3a+2y^{3a - (-2)} = y^{3a+2}.
3
Equate the exponents of like bases to find the values of aa and bb.
b=1b = 1 and a=4a = 4
Comparing x1y3a+2x^1 y^{3a+2} to xby14x^b y^{14} gives b=1b = 1 and 3a+2=143a + 2 = 14. Solving 3a+2=143a + 2 = 14 yields 3a=123a = 12, which simplifies to a=4a = 4.
4
Sum the values of aa and bb.
55
Adding aa and bb yields 4+1=54 + 1 = 5.

Anahtar Kavram

Applying properties of exponents, including the power of a product, power of a power, and quotient rules, to simplify algebraic expressions.
Tahmini Süre:1m 30s
Soru 225Soru

In the standard (x,y)(x, y) coordinate plane, a line segment has endpoints at (3,4)(3, -4) and (9,8)(9, 8). What is the yy-coordinate of the midpoint of this line segment?

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Cevap: 2

Cevap

The yy-coordinate of the midpoint is 22.
The yy-coordinate of the midpoint is calculated by finding the average of the yy-coordinates of the endpoints: y1+y22\frac{y_1 + y_2}{2}. Substituting y1=4y_1 = -4 and y2=8y_2 = 8 gives 4+82=42=2\frac{-4 + 8}{2} = \frac{4}{2} = 2.

Adım Adım Çözüm

1
Identify the yy-coordinates of the two given endpoints (3,4)(3, -4) and (9,8)(9, 8).
y1=4y_1 = -4 and y2=8y_2 = 8
The midpoint formula relies on the coordinates of the endpoints.
2
Calculate the average of the yy-coordinates using the formula ym=y1+y22y_m = \frac{y_1 + y_2}{2}.
ym=4+82=2y_m = \frac{-4 + 8}{2} = 2
The yy-coordinate of a midpoint is the arithmetic mean of the yy-coordinates of the endpoints.

Anahtar Kavram

Midpoint Formula
Tahmini Süre:45s
Soru 226Soru

A botanist is monitoring the heights of two bamboo plants. Plant A is 2020 inches tall and grows at a constant rate of 1.51.5 inches per day. Plant B is 1212 inches tall and grows at a constant rate of 2.52.5 inches per day. After how many days will Plant B be exactly 66 inches taller than Plant A?

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Cevap: 14

Cevap

The correct answer is 14 days.
The correct answer is 1414 days because when we translate the relationship, we get the equation 12+2.5d=20+1.5d+612 + 2.5d = 20 + 1.5d + 6. Solving this equation yields d=14d = 14.

Adım Adım Çözüm

1
Define the variable dd as the number of days and write the expressions for the heights of both plants.
Plant A's height is 20+1.5d20 + 1.5d inches, and Plant B's height is 12+2.5d12 + 2.5d inches.
To represent the growth of each plant algebraically over time.
2
Set up an equation representing that Plant B's height is 66 inches more than Plant A's height.
12+2.5d=(20+1.5d)+612 + 2.5d = (20 + 1.5d) + 6
To translate the verbal relationship into a mathematical equation.
3
Simplify the equation and solve for dd.
d=14d = 14
To find the number of days that satisfies the given condition.

Anahtar Kavram

Translating and Solving Algebraic Word Problems
Tahmini Süre:1m 15s
Soru 227Soru

A hyperbola in the standard (x,y)(x, y) coordinate plane is defined by the equation 9x216y236x32y124=09x^2 - 16y^2 - 36x - 32y - 124 = 0. One of the foci of this hyperbola is located at the point (f,1)(f, -1), where f>0f > 0. What is the value of ff?

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Cevap: 7

Cevap

7
Completing the square transforms the equation into the standard form of a horizontal hyperbola, (x2)216(y+1)29=1\frac{(x - 2)^2}{16} - \frac{(y + 1)^2}{9} = 1, which has its center at (2,1)(2, -1) with a2=16a^2 = 16 and b2=9b^2 = 9. The distance to the foci is c=a2+b2=16+9=5c = \sqrt{a^2 + b^2} = \sqrt{16 + 9} = 5. Since the transverse axis is horizontal, the foci are located at (2±5,1)(2 \pm 5, -1), which are (3,1)(-3, -1) and (7,1)(7, -1). Given the constraint that f>0f > 0, the positive x-coordinate of the focus is 7.

Adım Adım Çözüm

1
Group the terms and prepare to complete the square.
9(x24x)16(y2+2y)=1249(x^2 - 4x) - 16(y^2 + 2y) = 124
Grouping the variables helps isolate the quadratic expressions for completing the square.
2
Complete the square for both the xx and yy terms.
9(x2)216(y+1)2=1449(x - 2)^2 - 16(y + 1)^2 = 144
To complete the square for x24xx^2 - 4x, add 4 inside the first parentheses, adding 9×4=369 \times 4 = 36 to the right side. To complete the square for y2+2yy^2 + 2y, add 1 inside the second parentheses, which subtracts 16×1=1616 \times 1 = 16 from the right side because of the leading negative coefficient. This leaves the right side as 124+3616=144124 + 36 - 16 = 144.
3
Divide both sides of the equation by the constant to find the standard form.
(x2)216(y+1)29=1\frac{(x - 2)^2}{16} - \frac{(y + 1)^2}{9} = 1
Dividing by 144 puts the equation in the standard horizontal hyperbola form: (xh)2a2(yk)2b2=1\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1.
4
Find the distance cc from the center to the foci.
c=5c = 5
For a hyperbola, the relationship between the semi-axes and the focal distance is c=a2+b2c = \sqrt{a^2 + b^2}. Substituting a2=16a^2 = 16 and b2=9b^2 = 9 gives c=16+9=5c = \sqrt{16 + 9} = 5.
5
Determine the coordinates of the foci and extract the value of ff.
f=7f = 7
The center of the hyperbola is (h,k)=(2,1)(h, k) = (2, -1). The foci are located at (h±c,k)=(2±5,1)(h \pm c, k) = (2 \pm 5, -1), which corresponds to the points (3,1)(-3, -1) and (7,1)(7, -1). Since the problem states f>0f > 0, the target focus must be (7,1)(7, -1), meaning f=7f = 7.

Anahtar Kavram

Converting a general hyperbola equation into standard form to calculate focal points
Soru 228Soru

A certain radioactive isotope decays according to the formula N(t)=N02t/8N(t) = N_0 \cdot 2^{-t/8}, where N0N_0 is the initial amount of the isotope and tt is the time in years. If a sample initially contains 120120 grams of the isotope, how many years will it take for the amount of the isotope to decay to 1515 grams?

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Cevap: 24

Cevap

24
The correct answer is 24 because substituting the initial value of 120 and the final value of 15 into the equation yields 15=1202t/815 = 120 \cdot 2^{-t/8}. Dividing both sides by 120 gives 18=2t/8\frac{1}{8} = 2^{-t/8}, which can be rewritten as 23=2t/82^{-3} = 2^{-t/8}. Setting the exponents equal to each other gives 3=t/8-3 = -t/8, and solving for tt yields 24.

Adım Adım Çözüm

1
Substitute the given values into the decay formula.
15=1202t/815 = 120 \cdot 2^{-t/8}
The initial amount N0N_0 is 120120 grams and the final amount N(t)N(t) is 1515 grams.
2
Isolate the exponential term.
18=2t/8\frac{1}{8} = 2^{-t/8}
Divide both sides by 120120. Since 15120\frac{15}{120} reduces to 18\frac{1}{8}, this isolates the base 22 term.
3
Write the fraction as a power with base 2.
23=2t/82^{-3} = 2^{-t/8}
Using exponent rules, 18=123=23\frac{1}{8} = \frac{1}{2^3} = 2^{-3}.
4
Equate the exponents and solve for tt.
t=24t = 24
Since the bases are equal, the exponents must be equal, so 3=t8-3 = -\frac{t}{8} which gives t=24t = 24.

Anahtar Kavram

Solving exponential equations using a common base.
Soru 229Soru

In the standard (x,y)(x, y) coordinate plane, the line defined by the equation 3x4y=243x - 4y = 24 intersects the xx-axis at a certain point. What is the xx-coordinate of this intersection point?

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Cevap: 8

Cevap

The xx-coordinate of the intersection point is 88.
To find the xx-intercept, set y=0y = 0 in the equation 3x4y=243x - 4y = 24. This simplifies to 3x=243x = 24. Dividing both sides of the equation by 33 gives x=8x = 8.

Adım Adım Çözüm

1
Set y=0y = 0 in the equation.
3x4(0)=24    3x=243x - 4(0) = 24 \implies 3x = 24
The intersection of any graph with the xx-axis occurs where the yy-coordinate is 00.
2
Solve for xx.
x=8x = 8
Dividing both sides of 3x=243x = 24 by 33 isolates the variable xx.

Anahtar Kavram

Finding the xx-intercept of a line by setting y=0y = 0

Alternatif Yöntem

Convert the standard form equation 3x4y=243x - 4y = 24 into slope-intercept form: 4y=3x+24    y=34x6-4y = -3x + 24 \implies y = \frac{3}{4}x - 6. To find the xx-intercept, set y=0y = 0 and solve the equation 0=34x6    6=34x    x=80 = \frac{3}{4}x - 6 \implies 6 = \frac{3}{4}x \implies x = 8.
Tahmini Süre:45s
Soru 230Soru

In PQR\triangle PQR, the measure of P\angle P is 5050^\circ, and the measure of the exterior angle at vertex QQ is 110110^\circ. If the bisector of PRQ\angle PRQ intersects side PQPQ at point SS, what is the measure, in degrees, of PRS\angle PRS?

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Cevap: 30

Cevap

The measure of PRS\angle PRS is 3030^\circ.
By using the linear pair relationship, the interior angle PQR\angle PQR is found to be 7070^\circ. Applying the triangle angle sum theorem, the third interior angle PRQ\angle PRQ is 180(50+70)=60180^\circ - (50^\circ + 70^\circ) = 60^\circ. The angle bisector RSRS divides this angle into two equal parts, resulting in a measure of 3030^\circ for PRS\angle PRS.

Adım Adım Çözüm

1
Find the interior angle at vertex QQ
PQR=70\angle PQR = 70^\circ
The interior and exterior angles at a vertex are supplementary, summing to 180180^\circ.
2
Find the measure of interior angle PRQ\angle PRQ
PRQ=60\angle PRQ = 60^\circ
The sum of the interior angles in any triangle is 180180^\circ.
3
Calculate the measure of the bisected angle PRS\angle PRS
PRS=30\angle PRS = 30^\circ
An angle bisector divides the angle into two equal measures.

Anahtar Kavram

Triangle Angle Sum Theorem and Exterior Angle Relationships
Tahmini Süre:1m 30s
Soru 231Soru

The daily revenue RR, in dollars, of a manufacturing company is modeled by the quadratic function R(x)=0.2x2+kx1,200R(x) = -0.2x^2 + kx - 1,200, where xx is the number of units produced and sold, and kk is a positive constant. If the maximum daily revenue the company can achieve is 800800 dollars, what is the value of kk?

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Cevap: 40

Cevap

The value of the constant kk is 40.
Setting the daily revenue function equal to 800 and rewriting it in standard form yields 0.2x2+kx2000=0-0.2x^2 + kx - 2000 = 0. For a quadratic equation to have exactly one real solution, which represents the maximum vertex of the parabola, the discriminant must be equal to 0. Setting the discriminant b24ac=0b^2 - 4ac = 0 gives k24(0.2)(2000)=0k^2 - 4(-0.2)(-2000) = 0, which simplifies to k21600=0k^2 - 1600 = 0. Solving for the positive constant kk gives k=40k = 40.

Adım Adım Çözüm

1
Set the revenue function equal to the maximum daily revenue of 800 dollars.
0.2x2+kx1,200=800-0.2x^2 + kx - 1,200 = 800
The maximum revenue is the highest point (vertex) on the parabola, where the line y=800y = 800 is tangent to the curve.
2
Convert the equation into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
0.2x2+kx2,000=0-0.2x^2 + kx - 2,000 = 0
Standard form is required to identify the coefficients aa, bb, and cc for the discriminant formula.
3
Set the discriminant equal to zero.
k24(0.2)(2,000)=0k^2 - 4(-0.2)(-2,000) = 0
Since the maximum daily revenue is achieved at exactly one point, the quadratic equation must have exactly one real solution, meaning its discriminant (b24acb^2 - 4ac) must be zero.
4
Solve for the positive constant kk.
k21,600=0    k=40k^2 - 1,600 = 0 \implies k = 40
Solving the equation yields k=±40k = \pm 40. Since the problem specifies that kk is a positive constant, we select k=40k = 40.

Anahtar Kavram

Using the discriminant of a quadratic equation to find the value of a parameter when there is exactly one real solution.
Tahmini Süre:2m 0s
Soru 232Soru

In the standard (x,y)(x, y) coordinate plane, a line segment ABAB has midpoint M(3,4)M(3, 4). If endpoint AA lies on the line y=2x7y = 2x - 7, and the length of segment ABAB is 1010 units, what is the product of all possible xx-coordinates of endpoint AA?

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Cevap: 21

Cevap

The product of all possible xx-coordinates of endpoint AA is 2121.
By determining that the distance from endpoint A(x,2x7)A(x, 2x - 7) to the midpoint M(3,4)M(3, 4) is half of the segment length ABAB (which is 55), we set up the distance formula equation: (x3)2+(2x11)2=25(x - 3)^2 + (2x - 11)^2 = 25. Simplifying this equation leads to the quadratic expression x210x+21=0x^2 - 10x + 21 = 0, which factors into (x3)(x7)=0(x - 3)(x - 7) = 0. The two possible xx-coordinates are 33 and 77. Multiplying these values yields the product 2121.

Adım Adım Çözüm

1
Find the distance between endpoint AA and midpoint MM.
The distance AMAM is 55.
Since MM is the midpoint of segment ABAB of length 1010, the distance from either endpoint to the midpoint is half of the total length: 10÷2=510 \div 2 = 5.
2
Express the coordinates of endpoint AA in terms of a single variable.
Endpoint AA is represented as (x,2x7)(x, 2x - 7).
Endpoint AA lies on the line y=2x7y = 2x - 7.
3
Apply the distance formula to find the relationship for xx.
(x3)2+(2x11)2=25(x - 3)^2 + (2x - 11)^2 = 25
The distance between A(x,2x7)A(x, 2x - 7) and M(3,4)M(3, 4) is 55, so the square of the distance is 52=255^2 = 25.
4
Simplify the quadratic equation.
x210x+21=0x^2 - 10x + 21 = 0
Expanding (x3)2+(2x11)2=25(x - 3)^2 + (2x - 11)^2 = 25 yields x26x+9+4x244x+121=25x^2 - 6x + 9 + 4x^2 - 44x + 121 = 25. Combining like terms gives 5x250x+130=255x^2 - 50x + 130 = 25. Subtracting 2525 from both sides gives 5x250x+105=05x^2 - 50x + 105 = 0. Dividing the entire equation by 55 yields x210x+21=0x^2 - 10x + 21 = 0.
5
Solve for the possible values of xx.
x=3x = 3 or x=7x = 7
Factoring the quadratic equation yields (x3)(x7)=0(x - 3)(x - 7) = 0, which gives the roots x=3x = 3 and x=7x = 7.
6
Calculate the product of the possible xx-coordinates.
21
The product of the two possible xx-coordinates is 3×7=213 \times 7 = 21.

Anahtar Kavram

Distance and Midpoint Formulas

Alternatif Yöntem

Instead of expanding the quadratic equation, one can use the geometric interpretation. The points AA are the intersections of the circle (x3)2+(y4)2=25(x - 3)^2 + (y - 4)^2 = 25 and the line y=2x7y = 2x - 7. Substituting y=2x7y = 2x - 7 directly into the circle equation and simplifying to x210x+21=0x^2 - 10x + 21 = 0 is the most direct approach.
Tahmini Süre:2m 0s
Soru 233Soru

In the standard (x,y)(x, y) coordinate plane, a line with a negative slope passes through the point (4,3)(4, 3) and has an xx-intercept that is twice its yy-intercept. If the equation of this line is written in the form Ax+By=CAx + By = C, where AA, BB, and CC are integers with no common factor greater than 1, and A>0A > 0, what is the value of CC?

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Cevap: 10

Cevap

The value of CC is 10.
The correct answer is 10. By setting the intercepts as (2b,0)(2b, 0) and (0,b)(0, b), we find the slope is m=12m = -\frac{1}{2}. Applying the point-slope formula with (4,3)(4, 3) gives y3=12(x4)y - 3 = -\frac{1}{2}(x - 4), which simplifies to y=12x+5y = -\frac{1}{2}x + 5. Converting this to standard form with A>0A > 0 yields x+2y=10x + 2y = 10, where C=10C = 10.

Adım Adım Çözüm

1
Represent the coordinates of the intercepts using a variable.
The yy-intercept is (0,b)(0, b) and the xx-intercept is (2b,0)(2b, 0).
The problem states that the xx-intercept is twice the yy-intercept.
2
Calculate the slope (mm) of the line using the intercepts.
m=b002b=12m = \frac{b - 0}{0 - 2b} = -\frac{1}{2}
The slope of a line passing through (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}.
3
Find the equation of the line using the point-slope form with point (4,3)(4, 3).
y3=12(x4)y=12x+5y - 3 = -\frac{1}{2}(x - 4) \Rightarrow y = -\frac{1}{2}x + 5
The point-slope form of a linear equation is yy1=m(xx1)y - y_1 = m(x - x_1).
4
Convert the equation to the standard form Ax+By=CAx + By = C.
x+2y=10x + 2y = 10
Multiplying the equation by 2 and moving the xx term to the left side results in integer coefficients where the coefficient of xx is positive (A=1>0A = 1 > 0).
5
Identify the value of CC from the standard form equation.
C=10C = 10
Comparing x+2y=10x + 2y = 10 with Ax+By=CAx + By = C shows that A=1A=1, B=2B=2, and C=10C=10. These integers share no common factors greater than 1.

Anahtar Kavram

Converting a linear equation to standard form using given geometric features and a coordinate point.

Alternatif Yöntem

Instead of using the point-slope formula, substitute the point (4,3)(4, 3) directly into the intercept form of a linear equation, xa+yb=1\frac{x}{a} + \frac{y}{b} = 1. Since a=2ba = 2b, this becomes 42b+3b=1\frac{4}{2b} + \frac{3}{b} = 1. Simplifying this gives 2b+3b=15b=1b=5\frac{2}{b} + \frac{3}{b} = 1 \Rightarrow \frac{5}{b} = 1 \Rightarrow b = 5. Thus, a=10a = 10. The intercept form is x10+y5=1\frac{x}{10} + \frac{y}{5} = 1. Multiplying by 10 to clear denominators gives x+2y=10x + 2y = 10, where C=10C = 10.
Tahmini Süre:1m 30s
Soru 234Soru

In the standard (x,y)(x, y) coordinate plane, the perpendicular bisector of the line segment with endpoints (1,2)(1, 2) and (5,10)(5, 10) intersects the curve y=x2y = x^2 at two points. If one of these intersection points lies in the second quadrant, what is the yy-coordinate of this point?

Cevabı ve açıklamayı göster

Cevap: 9

Cevap

The correct answer is 99.
The perpendicular bisector of the segment connecting (1,2)(1, 2) and (5,10)(5, 10) passes through their midpoint (3,6)(3, 6) and has a slope of 12-\frac{1}{2} (the negative reciprocal of 22). The equation of this line is x+2y=15x + 2y = 15. Substituting y=x2y = x^2 yields the quadratic equation 2x2+x15=02x^2 + x - 15 = 0, which factors into (2x5)(x+3)=0(2x - 5)(x + 3) = 0. The solution in the second quadrant corresponds to the negative xx-coordinate, x=3x = -3. Squaring this value gives a yy-coordinate of 99.

Adım Adım Çözüm

1
Find the midpoint of the segment with endpoints (1,2)(1, 2) and (5,10)(5, 10).
The midpoint is (3,6)(3, 6).
The perpendicular bisector of a segment passes through its midpoint.
2
Calculate the slope of the segment and the slope of the perpendicular bisector.
The slope of the segment is 22, and the perpendicular slope is 12-\frac{1}{2}.
Perpendicular lines have slopes that are negative reciprocals of each other.
3
Write the equation of the perpendicular bisector.
The equation of the line is x+2y=15x + 2y = 15.
Using the point-slope form with point (3,6)(3, 6) and slope 12-\frac{1}{2} yields y6=12(x3)y - 6 = -\frac{1}{2}(x - 3), which simplifies to x+2y=15x + 2y = 15.
4
Find the intersection points of the line and the curve y=x2y = x^2.
The xx-coordinates of the intersection points are 2.52.5 and 3-3.
Substituting y=x2y = x^2 into x+2y=15x + 2y = 15 gives the quadratic equation 2x2+x15=02x^2 + x - 15 = 0, which factors as (2x5)(x+3)=0(2x - 5)(x + 3) = 0.
5
Identify the point in the second quadrant and find its yy-coordinate.
The point is (3,9)(-3, 9), so the yy-coordinate is 99.
A point in the second quadrant must have a negative xx-coordinate (x=3x = -3) and a positive yy-coordinate (y=9y = 9).

Anahtar Kavram

Perpendicular bisectors and systems of linear-quadratic equations in coordinate geometry
Soru 235Soru

A delivery drone starts a flight with a battery charge of 95%95\%. The battery charge decreases at a constant rate of 1.5%1.5\% per minute of flight time. During the flight, the drone lands on a charging station for 1515 minutes, during which its battery charge increases at a constant rate of 2.4%2.4\% per minute. After this charging period, the drone resumes its flight. If the drone ends its flight with a battery charge of 65%65\%, and its total flight time (excluding the time spent charging) was tt minutes, what is the value of tt?

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Cevap: 44

Cevap

44
The correct answer is found by setting up a linear equation that models the drone's battery charge changes. The drone starts with 95%95\% battery, loses 1.5%1.5\% per minute for tt minutes, and gains 36%36\% from charging (15×2.4%15 \times 2.4\%). Setting this equal to the final charge of 65%65\% gives the equation 951.5t+36=6595 - 1.5t + 36 = 65. Simplifying gives 1311.5t=65131 - 1.5t = 65, which leads to 1.5t=66-1.5t = -66 and t=44t = 44.

Adım Adım Çözüm

1
Calculate the total percentage of battery charge gained while charging.
36%36\%
The drone charges at a rate of 2.4%2.4\% per minute for 1515 minutes, so the total gain is 15×2.4%=36%15 \times 2.4\% = 36\%.
2
Set up a linear equation for the final battery charge.
951.5t+36=6595 - 1.5t + 36 = 65
The final battery charge (65%65\%) is the initial charge (95%95\%) minus the battery consumed during flight (1.5%1.5\% per minute for tt minutes) plus the charge gained (36%36\%).
3
Combine constant terms on the left side of the equation.
1311.5t=65131 - 1.5t = 65
Adding 9595 and 3636 simplifies the expression on the left side.
4
Isolate the variable term by subtracting 131131 from both sides.
1.5t=66-1.5t = -66
Subtracting 131131 from both sides leaves only the variable term on the left.
5
Divide both sides by 1.5-1.5 to solve for tt.
t=44t = 44
Dividing 66-66 by 1.5-1.5 isolates tt to find the total flight time.

Anahtar Kavram

Solving linear equations in a real-world context by setting up an algebraic equation.
Tahmini Süre:1m 30s
Soru 236Soru

In PQR\triangle PQR, the lengths of sides PQPQ and PRPR are equal. If the measure of P\angle P is 7070^\circ, what is the measure, in degrees, of Q\angle Q?

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Cevap: 55

Cevap

The measure of Q\angle Q is 5555^\circ.
Since PQ=PRPQ = PR, PQR\triangle PQR is an isosceles triangle with base QRQR, which means the base angles Q\angle Q and R\angle R have equal measures. The sum of the interior angles of a triangle is 180180^\circ. Setting up the equation gives 70+2(Q)=18070^\circ + 2(\angle Q) = 180^\circ. Subtracting 7070^\circ from both sides yields 2(Q)=1102(\angle Q) = 110^\circ. Dividing by 2, we find that the measure of Q\angle Q is 5555^\circ.

Adım Adım Çözüm

1
Identify the properties of the given triangle.
PQR\triangle PQR is an isosceles triangle with base QRQR and base angles Q=R\angle Q = \angle R.
A triangle with two equal sides is isosceles, and the angles opposite those sides are equal in measure.
2
Apply the triangle angle sum theorem.
P+Q+R=180\angle P + \angle Q + \angle R = 180^\circ
The sum of the measures of the interior angles of any triangle is always 180180^\circ.
3
Substitute the known values and solve for Q\angle Q.
70+2(Q)=180    2(Q)=110    Q=5570^\circ + 2(\angle Q) = 180^\circ \implies 2(\angle Q) = 110^\circ \implies \angle Q = 55^\circ
Substituting P=70\angle P = 70^\circ and R=Q\angle R = \angle Q allows us to solve the linear equation for the unknown angle measure.

Anahtar Kavram

Isosceles Triangle Properties and Triangle Angle Sum Theorem
Tahmini Süre:45s
Soru 237Soru

For the functions f(x)=4x17f(x) = |4x - 17| and g(x)=32xg(x) = 3 - 2x, what is the value of f(g(5))f(g(5))?

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Cevap: 45

Cevap

The correct answer is 45.
Evaluating the inner function first yields g(5)=32(5)=7g(5) = 3 - 2(5) = -7. Substituting this result into the outer function gives f(7)=4(7)17=45=45f(-7) = |4(-7) - 17| = |-45| = 45.

Adım Adım Çözüm

1
Evaluate the inner function g(5)g(5)
g(5)=32(5)=7g(5) = 3 - 2(5) = -7
In a composite function of the form f(g(x))f(g(x)), the inner function g(x)g(x) must be evaluated first at the given input.
2
Evaluate the outer function f(x)f(x) at the output of the inner function
f(7)=4(7)17=2817=45=45f(-7) = |4(-7) - 17| = |-28 - 17| = |-45| = 45
The output of the inner function, 7-7, becomes the input for the outer function f(x)f(x).

Anahtar Kavram

Function Evaluation and Composition
Soru 238Soru

On a coordinate map, two tracking stations are located at the points (2,3)(2, -3) and (6,12)(-6, 12). What is the straight-line distance, in map units, between the two stations?

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Cevap: 17

Cevap

The straight-line distance between the two stations is 17 map units.
Applying the coordinate distance formula to the coordinates (2,3)(2, -3) and (6,12)(-6, 12) yields a distance of (62)2+(12(3))2=(8)2+152=64+225=289=17\sqrt{(-6 - 2)^2 + (12 - (-3))^2} = \sqrt{(-8)^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17.

Adım Adım Çözüm

1
Identify coordinates of the two stations.
(x1,y1)=(2,3)(x_1, y_1) = (2, -3) and (x2,y2)=(6,12)(x_2, y_2) = (-6, 12)
Defining the coordinate variables is necessary to apply the formula correctly.
2
Apply the distance formula.
d=(62)2+(12(3))2d = \sqrt{(-6 - 2)^2 + (12 - (-3))^2}
The distance formula calculates the straight-line distance between two coordinates.
3
Simplify the arithmetic terms.
d=(8)2+152=64+225=289d = \sqrt{(-8)^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289}
Simplify the differences and squares under the radical.
4
Calculate the final square root.
d=17d = 17
The square root of 289 is 17.

Anahtar Kavram

Distance Formula
Soru 239Soru

A hyperbola is defined by the equation 9x24y236x8y4=09x^2 - 4y^2 - 36x - 8y - 4 = 0 in the standard (x,y)(x, y) coordinate plane. What is the slope of the asymptote of this hyperbola that has a positive slope?

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Cevap: 1.5

Cevap

The positive slope of the asymptotes is 1.5.
By converting the general form of the hyperbola equation into standard form, we determine that it is a horizontal hyperbola with a=2a = 2 and b=3b = 3. The slopes of the asymptotes for a horizontal hyperbola are ±ba\pm \frac{b}{a}, making the positive slope equal to 32=1.5\frac{3}{2} = 1.5.

Adım Adım Çözüm

1
Group x-terms and y-terms, and move the constant to the other side.
9(x24x)4(y2+2y)=49(x^2 - 4x) - 4(y^2 + 2y) = 4
This prepares the equation for completing the square.
2
Complete the square for the quadratic expressions in x and y.
9(x2)24(y+1)2=369(x - 2)^2 - 4(y + 1)^2 = 36
Completing the square yields 9[(x2)24]4[(y+1)21]=4    9(x2)2364(y+1)2+4=49[(x-2)^2 - 4] - 4[(y+1)^2 - 1] = 4 \implies 9(x-2)^2 - 36 - 4(y+1)^2 + 4 = 4.
3
Divide both sides by 36 to format the equation in standard hyperbola form.
(x2)24(y+1)29=1\frac{(x-2)^2}{4} - \frac{(y+1)^2}{9} = 1
The standard form of a horizontal hyperbola centered at (h,k)(h, k) is (xh)2a2(yk)2b2=1\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1.
4
Identify the values of a and b from the denominators.
a=2a = 2 and b=3b = 3
Since a2=4a^2 = 4 and b2=9b^2 = 9, taking the square roots gives a=2a = 2 and b=3b = 3.
5
Determine the positive slope of the asymptotes using the formula for a horizontal hyperbola.
m=ba=1.5m = \frac{b}{a} = 1.5
The asymptotes for a horizontal hyperbola are given by yk=±ba(xh)y - k = \pm \frac{b}{a}(x - h), so the positive slope is ba=32=1.5\frac{b}{a} = \frac{3}{2} = 1.5.

Anahtar Kavram

Rewriting a hyperbola equation from general form to standard form to find asymptote equations.
Tahmini Süre:1m 30s
Soru 240Soru

A line segment in the standard (x,y)(x, y) coordinate plane has endpoints at (1,a)(1, a) and (5,a2)(5, a^2). The perpendicular bisector of this segment is parallel to the line defined by the equation x+3y=6x + 3y = 6. What is the sum of all possible values of the constant aa?

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Cevap: 1

Cevap

The sum of all possible values of the constant aa is 1.
To find the sum of all possible values of the constant aa, we first find the slope of the line x+3y=6x + 3y = 6 by writing it in slope-intercept form: y=13x+2y = -\frac{1}{3}x + 2. Since the perpendicular bisector is parallel to this line, its slope is also 13-\frac{1}{3}. The line segment is perpendicular to its perpendicular bisector, so the slope of the line segment is the negative reciprocal of 13-\frac{1}{3}, which is 33. Setting the slope of the segment a2a51\frac{a^2 - a}{5 - 1} equal to 33 gives the equation a2a4=3\frac{a^2 - a}{4} = 3, which simplifies to a2a12=0a^2 - a - 12 = 0. Solving this quadratic equation yields (a4)(a+3)=0(a - 4)(a + 3) = 0, giving the values a=4a = 4 and a=3a = -3. The sum of these possible values is 4+(3)=14 + (-3) = 1.

Adım Adım Çözüm

1
Find the slope of the given line x+3y=6x + 3y = 6.
The slope is 13-\frac{1}{3}.
Rewriting the equation in slope-intercept form (y=mx+by = mx + b) gives y=13x+2y = -\frac{1}{3}x + 2, showing the slope is 13-\frac{1}{3}.
2
Determine the slope of the line segment.
The slope is 3.
The line segment is perpendicular to its perpendicular bisector. Because the perpendicular bisector is parallel to the reference line, its slope is also 13-\frac{1}{3}. The line segment's slope is the negative reciprocal of 13-\frac{1}{3}, which is 33.
3
Write the slope of the segment in terms of aa and set it equal to 3.
a2a4=3\frac{a^2 - a}{4} = 3
Using the slope formula with endpoints (1,a)(1, a) and (5,a2)(5, a^2) gives the expression a2a51\frac{a^2 - a}{5 - 1}.
4
Solve the quadratic equation for aa.
a=4a = 4 or a=3a = -3
Multiplying both sides by 4 yields a2a=12a^2 - a = 12, which simplifies to the quadratic a2a12=0a^2 - a - 12 = 0. Factoring gives (a4)(a+3)=0(a - 4)(a + 3) = 0.
5
Sum all possible values of aa.
1
The sum of the values is 4+(3)=14 + (-3) = 1.

Anahtar Kavram

Understanding that parallel lines have equal slopes, perpendicular lines have slopes that are negative reciprocals of each other, and applying the slope formula to solve for coordinate variables.
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