Algebra

239 soru

Soru 141Soru

Find the product matrix ABAB given that A=(1203)A = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix} and B=(4512)B = \begin{pmatrix} 4 & 5 \\ 1 & 2 \end{pmatrix}.

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Cevap: (6936)\begin{pmatrix} 6 & 9 \\ 3 & 6 \end{pmatrix}

Cevap

(6936)\begin{pmatrix} 6 & 9 \\ 3 & 6 \end{pmatrix}
The matrix product ABAB is found by performing row-by-column multiplication. Computing row 1 gives (1)(4)+(2)(1)=6(1)(4)+(2)(1) = 6 and (1)(5)+(2)(2)=9(1)(5)+(2)(2) = 9. Computing row 2 gives (0)(4)+(3)(1)=3(0)(4)+(3)(1) = 3 and (0)(5)+(3)(2)=6(0)(5)+(3)(2) = 6. Thus, the resulting matrix is (6936)\begin{pmatrix} 6 & 9 \\ 3 & 6 \end{pmatrix}.

Adım Adım Çözüm

1
Set up row-by-column matrix multiplication formula for 2×22 \times 2 matrices.
The entry at row ii, column jj of product ABAB is the dot product of row ii of AA and column jj of BB.
Matrix multiplication requires combining rows of the left matrix with columns of the right matrix.
2
Calculate row 1 entries of the product matrix.
First element: (1)(4)+(2)(1)=4+2=6(1)(4) + (2)(1) = 4 + 2 = 6. Second element: (1)(5)+(2)(2)=5+4=9(1)(5) + (2)(2) = 5 + 4 = 9.
Multiply elements of row 1 of AA by corresponding elements of columns 1 and 2 of BB and sum them.
3
Calculate row 2 entries of the product matrix.
Third element: (0)(4)+(3)(1)=0+3=3(0)(4) + (3)(1) = 0 + 3 = 3. Fourth element: (0)(5)+(3)(2)=0+6=6(0)(5) + (3)(2) = 0 + 6 = 6.
Multiply elements of row 2 of AA by corresponding elements of columns 1 and 2 of BB and sum them.
4
Assemble the product matrix.
AB=(6936)AB = \begin{pmatrix} 6 & 9 \\ 3 & 6 \end{pmatrix}.
Combine all computed row and column entries into a single 2×22 \times 2 matrix.

Anahtar Kavram

Matrix Multiplication (2×22 \times 2 Matrix Product)
Soru 142Soru

If log2(x+3)+log24=5\log_2 (x + 3) + \log_2 4 = 5, what is the value of xx?

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Cevap: 55

Cevap

The value of xx is 55.
Using the logarithmic addition property logb(M)+logb(N)=logb(MN)\log_b (M) + \log_b (N) = \log_b (M \cdot N), the equation log2(x+3)+log24=5\log_2(x+3) + \log_2 4 = 5 becomes log2[4(x+3)]=5\log_2[4(x+3)] = 5. Converting to index form gives 4(x+3)=25=324(x+3) = 2^5 = 32. Dividing by 44 yields x+3=8x + 3 = 8, so x=5x = 5.

Adım Adım Çözüm

1
Apply the product rule of logarithms: logbA+logbB=logb(AB)\log_b A + \log_b B = \log_b (A \cdot B)
\log_2 [4(x + 3)] = 5
Logarithms with the same base that are added together can be combined by multiplying their arguments.
2
Convert the logarithmic equation into its equivalent exponential form
4(x + 3) = 2^5 = 32
If logbY=c\log_b Y = c, then Y=bcY = b^c.
3
Divide both sides by 4 and solve for xx
x + 3 = 8 \implies x = 5
Linear algebraic simplification.

Anahtar Kavram

Product Law of Logarithms and Log-to-Exponential Conversion
Soru 143Soru

What is the simplified form of the expression 122+35\frac{12}{\sqrt{2} + \sqrt{3} - \sqrt{5}}?

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Cevap: 32+23+303\sqrt{2} + 2\sqrt{3} + \sqrt{30}

Cevap

32+23+303\sqrt{2} + 2\sqrt{3} + \sqrt{30}
By grouping the denominator as (2+3)5(\sqrt{2}+\sqrt{3}) - \sqrt{5} and multiplying by its conjugate (2+3)+5(\sqrt{2}+\sqrt{3}) + \sqrt{5}, the denominator reduces to 262\sqrt{6}. Multiplying the resulting fraction by 6/6\sqrt{6}/\sqrt{6} yields 12+18+30\sqrt{12} + \sqrt{18} + \sqrt{30}, which simplifies directly to 32+23+303\sqrt{2} + 2\sqrt{3} + \sqrt{30}.

Adım Adım Çözüm

1
Group the terms in the denominator as ((2+3)5)((\sqrt{2} + \sqrt{3}) - \sqrt{5}) and multiply the numerator and denominator by its conjugate ((2+3)+5)((\sqrt{2} + \sqrt{3}) + \sqrt{5}).
The fraction becomes 12((2+3)+5)((2+3)5)((2+3)+5)\frac{12((\sqrt{2} + \sqrt{3}) + \sqrt{5})}{((\sqrt{2} + \sqrt{3}) - \sqrt{5})((\sqrt{2} + \sqrt{3}) + \sqrt{5})}.
Applying the difference of two squares to eliminate the outer radical.
2
Expand the denominator using (ab)(a+b)=a2b2(a-b)(a+b) = a^2 - b^2.
Denominator = (2+3)2(5)2=(2+26+3)5=26(\sqrt{2} + \sqrt{3})^2 - (\sqrt{5})^2 = (2 + 2\sqrt{6} + 3) - 5 = 2\sqrt{6}.
Simplifying the algebraic square of a binomial surd.
3
Divide the numerator by the constant factor of the denominator.
\frac{12(\sqrt{2} + \sqrt{3} + \sqrt{5})}{2\sqrt{6}} = \frac{6(\sqrt{2} + \sqrt{3} + \sqrt{5})}{\sqrt{6}}.
Simplifying numerical coefficients before further rationalization.
4
Rationalize the remaining monomial radical in the denominator by multiplying numerator and denominator by 6\sqrt{6}.
\frac{6(\sqrt{12} + \sqrt{18} + \sqrt{30})}{6} = \sqrt{12} + \sqrt{18} + \sqrt{30}.
Eliminating 6\sqrt{6} from the denominator.
5
Simplify each radical to its simplest surd form.
\sqrt{12} = 2\sqrt{3}, \quad \sqrt{18} = 3\sqrt{2}, \quad \text{so } \sqrt{12} + \sqrt{18} + \sqrt{30} = 3\sqrt{2} + 2\sqrt{3} + \sqrt{30}.
Factoring out perfect square components from radical terms.

Anahtar Kavram

Rationalisation of trinomial surd denominators using repeated conjugate multiplication
Tahmini Süre:2m 0s
Soru 144Soru

What is the sum of all real values of xx that satisfy the logarithmic equation log3(9x+81)x=log330\log_3(9^x + 81) - x = \log_3 30?

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Cevap: 4

Cevap

The sum of all real values of xx satisfying the equation is 4.
Using logarithmic properties, log3(9x+81)x=log330\log_3(9^x + 81) - x = \log_3 30 simplifies to 9x+813x=30\frac{9^x + 81}{3^x} = 30. Substituting u=3xu = 3^x produces the quadratic equation u230u+81=0u^2 - 30u + 81 = 0, which yields u=3u = 3 and u=27u = 27. Solving 3x=33^x = 3 and 3x=273^x = 27 gives x=1x = 1 and x=3x = 3. Their sum is 1+3=41 + 3 = 4.

Adım Adım Çözüm

1
Express xx as log3(3x)\log_3(3^x) and apply the logarithm quotient rule.
\log_3\left(\frac{9^x + 81}{3^x}\right) = \log_3 30
Using logbAlogbB=logb(AB)\log_b A - \log_b B = \log_b\left(\frac{A}{B}\right) combines terms into a single logarithm.
2
Equate the logarithmic arguments and split the fraction.
\frac{9^x + 81}{3^x} = 30 \implies 3^x + \frac{81}{3^x} = 30
Logarithmic equality with identical bases implies equality of their arguments.
3
Substitute u=3xu = 3^x to form and solve a quadratic equation.
u + \frac{81}{u} = 30 \implies u^2 - 30u + 81 = 0 \implies (u - 3)(u - 27) = 0 \implies u = 3 \text{ or } u = 27
Since 9x=(3x)2=u29^x = (3^x)^2 = u^2, substituting u=3xu = 3^x converts the equation into a standard quadratic equation.
4
Solve for xx from 3x=u3^x = u and calculate the sum of the solutions.
3^x = 3 \implies x = 1; \quad 3^x = 27 \implies x = 3. \quad \text{Sum} = 1 + 3 = 4
Matching exponential bases gives x=1x = 1 and x=3x = 3.

Anahtar Kavram

Logarithmic quotient identity and quadratic substitution for exponential equations
Tahmini Süre:2m 0s
Soru 145Soru

An infinite geometric progression of positive terms has a sum to infinity of 1616, and the sum of its first two terms is 1212. An arithmetic progression has its first term equal to the first term of this geometric progression, and its 5th5^{\text{th}} term equal to the sum to infinity of the geometric progression. Calculate the 10th10^{\text{th}} term of the arithmetic progression.

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Cevap: 26

Cevap

The 10th term of the arithmetic progression is 26.
By solving the geometric progression system, we find the common ratio r=12r = \frac{1}{2} and first term a=8a = 8. Using a=8a = 8 as the first term of the arithmetic progression and setting its 5th5^{\text{th}} term A5=16A_5 = 16, we determine the common difference d=2d = 2. Calculating A10=8+9(2)A_{10} = 8 + 9(2) yields 2626.

Adım Adım Çözüm

1
Formulate equations for the geometric progression using the sum to infinity and sum of the first two terms
a=16(1r)a = 16(1 - r) and a(1+r)=12a(1 + r) = 12
The standard formula for the sum to infinity of a GP is S=a1rS_\infty = \frac{a}{1-r} and the sum of the first two terms is S2=a+ar=a(1+r)S_2 = a + ar = a(1+r).
2
Solve for the common ratio rr and first term aa of the geometric progression
r=0.5r = 0.5 and a=8a = 8
Substituting a=16(1r)a = 16(1-r) yields 16(1r2)=12    r2=14    r=1216(1-r^2) = 12 \implies r^2 = \frac{1}{4} \implies r = \frac{1}{2}. Then a=16(10.5)=8a = 16(1 - 0.5) = 8.
3
Determine the common difference dd of the arithmetic progression
d=2d = 2
The first term of the AP is A1=a=8A_1 = a = 8 and the 5th term is A5=S=16A_5 = S_\infty = 16. Using A5=A1+4d    8+4d=16    d=2A_5 = A_1 + 4d \implies 8 + 4d = 16 \implies d = 2.
4
Calculate the 10th term of the arithmetic progression
A10=26A_{10} = 26
Using the AP nthn^{\text{th}} term formula An=A1+(n1)dA_n = A_1 + (n-1)d: A10=8+9(2)=26A_{10} = 8 + 9(2) = 26.

Anahtar Kavram

Combining geometric progression parameters (sum to infinity and sum of terms) with arithmetic progression term formulas
Soru 146Soru

If log102=p\log_{10} 2 = p and log103=q\log_{10} 3 = q, which of the following expressions represents log1018\log_{10} 18 in terms of pp and qq?

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Cevap: p+2qp + 2q

Cevap

The expression for log1018\log_{10} 18 in terms of pp and qq is p+2qp + 2q.
Expanding log1018\log_{10} 18 as log10(2×32)\log_{10}(2 \times 3^2) gives log102+2log103\log_{10} 2 + 2\log_{10} 3, which simplifies directly to p+2qp + 2q.

Adım Adım Çözüm

1
Factorize 18 into prime factors.
18=2×3218 = 2 \times 3^2
Decomposing 18 into prime factors allows the application of the given logarithm values for 2 and 3.
2
Apply the product rule of logarithms: log10(a×b)=log10a+log10b\log_{10}(a \times b) = \log_{10} a + \log_{10} b.
log1018=log10(2×32)=log102+log10(32)\log_{10} 18 = \log_{10}(2 \times 3^2) = \log_{10} 2 + \log_{10}(3^2)
The logarithm of a product equals the sum of the logarithms of individual factors.
3
Apply the power rule of logarithms: log10(bn)=nlog10b\log_{10}(b^n) = n \log_{10} b.
log10(32)=2log103\log_{10}(3^2) = 2 \log_{10} 3
The logarithm of a power expression allows bringing the exponent to the front as a multiplier.
4
Substitute the defined values p=log102p = \log_{10} 2 and q=log103q = \log_{10} 3.
log1018=p+2q\log_{10} 18 = p + 2q
Replaces logarithmic terms with their algebraic representations.

Anahtar Kavram

Logarithm Expansion Laws (Product and Power Rules)
Soru 147Soru

Let the universal set be U={xZ:1x12}\mathcal{U} = \{x \in \mathbb{Z} : 1 \le x \le 12\}. Consider the subsets P={x:x is a factor of 12}P = \{x : x \text{ is a factor of } 12\} and Q={x:x is an even number,1x12}Q = \{x : x \text{ is an even number}, 1 \le x \le 12\}. Which of the following sets represents (PQ)(P \cup Q)'?

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Cevap: {5,7,9,11}\{5, 7, 9, 11\}

Cevap

{5,7,9,11}\{5, 7, 9, 11\}
The set of factors of 12 within the given range is {1, 2, 3, 4, 6, 12} and the set of even numbers up to 12 is {2, 4, 6, 8, 10, 12}. Their union contains all numbers that are either even or factors of 12, which is {1, 2, 3, 4, 6, 8, 10, 12}. Subtracting this union from the universal set {1, 2, ..., 12} yields {5, 7, 9, 11}.

Adım Adım Çözüm

1
List all elements of the universal set and the subsets PP and QQ.
U={1,2,3,4,5,6,7,8,9,10,11,12}\mathcal{U} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}, P={1,2,3,4,6,12}P = \{1, 2, 3, 4, 6, 12\}, and Q={2,4,6,8,10,12}Q = \{2, 4, 6, 8, 10, 12\}.
Identify the element listings explicitly from the given definitions.
2
Find the union of sets PP and QQ, denoted by PQP \cup Q.
PQ={1,2,3,4,6,8,10,12}P \cup Q = \{1, 2, 3, 4, 6, 8, 10, 12\}.
Combine all unique elements belonging to either PP, QQ, or both.
3
Find the complement (PQ)(P \cup Q)' relative to the universal set U\mathcal{U}.
(PQ)=U(PQ)={5,7,9,11}(P \cup Q)' = \mathcal{U} \setminus (P \cup Q) = \{5, 7, 9, 11\}.
Select all elements in U\mathcal{U} that do not appear in PQP \cup Q.

Anahtar Kavram

Complement of a Set Union
Tahmini Süre:50s
Soru 148Soru

The 1st1^{\text{st}}, 2nd2^{\text{nd}}, and 5th5^{\text{th}} terms of an arithmetic progression (A.P.) with a non-zero common difference form three consecutive terms of a geometric progression (G.P.). If the sum of the first 44 terms of the A.P. is 4040, what is the 5th5^{\text{th}} term of the G.P.?

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Cevap: 4052\frac{405}{2}

Cevap

4052\frac{405}{2}
By setting up the geometric mean property (a+d)2=a(a+4d)(a+d)^2 = a(a+4d), we find d=2ad = 2a, which establishes that the G.P. has a common ratio r=3r = 3. Substituting d=2ad = 2a into the A.P. sum formula S4=2[2a+3d]=16a=40S_4 = 2[2a + 3d] = 16a = 40 gives a=52a = \frac{5}{2}. Finally, evaluating the 5th5^{\text{th}} term of the G.P. using ar4=52×34a r^4 = \frac{5}{2} \times 3^4 yields 4052\frac{405}{2}.

Adım Adım Çözüm

1
Express the given A.P. terms in terms of first term aa and common difference dd, and set up the G.P. condition.
The terms are T1=aT_1 = a, T2=a+dT_2 = a + d, and T5=a+4dT_5 = a + 4d. Since they form a G.P., (a+d)2=a(a+4d)(a + d)^2 = a(a + 4d).
Three terms x,y,zx, y, z form a G.P. if y2=xzy^2 = xz.
2
Solve for the relationship between dd and aa.
a2+2ad+d2=a2+4ad    d2=2ad    d=2aa^2 + 2ad + d^2 = a^2 + 4ad \implies d^2 = 2ad \implies d = 2a (since d0d \neq 0).
Expanding and simplifying the equation yields the ratio of dd to aa.
3
Determine the common ratio rr of the G.P.
r=T2T1=a+da=a+2aa=3r = \frac{T_2}{T_1} = \frac{a + d}{a} = \frac{a + 2a}{a} = 3.
The common ratio is the quotient of consecutive terms of the G.P.
4
Use the sum of the first 44 terms of the A.P. to find aa.
S4=42[2a+(41)d]=2[2a+3(2a)]=16a=40    a=4016=52S_4 = \frac{4}{2}[2a + (4-1)d] = 2[2a + 3(2a)] = 16a = 40 \implies a = \frac{40}{16} = \frac{5}{2}.
Applying Sn=n2[2a+(n1)d]S_n = \frac{n}{2}[2a + (n-1)d] with S4=40S_4 = 40 allows solving for aa.
5
Calculate the 5th5^{\text{th}} term of the G.P.
G5=g1r51=ar4=5234=5281=4052G_5 = g_1 \cdot r^{5-1} = a \cdot r^4 = \frac{5}{2} \cdot 3^4 = \frac{5}{2} \cdot 81 = \frac{405}{2}.
The nthn^{\text{th}} term of a G.P. is gn=g1rn1g_n = g_1 r^{n-1}.

Anahtar Kavram

Connecting Arithmetic and Geometric Progressions using term definitions and sum formulas.
Soru 149Soru

If 3+232323+2=k6\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} - \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} = k\sqrt{6}, find the value of the integer kk.

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Cevap: 4

Cevap

The value of the integer kk is 4.
Combining the fractions over the common denominator (32)(3+2)=32=1(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2}) = 3-2 = 1 yields a numerator of (3+2+26)(3+226)=46(3+2+2\sqrt{6}) - (3+2-2\sqrt{6}) = 4\sqrt{6}. Thus, k6=46k\sqrt{6} = 4\sqrt{6}, which gives k=4k = 4.

Adım Adım Çözüm

1
Combine the fractions using their common denominator
\frac{(\sqrt{3} + \sqrt{2})^2 - (\sqrt{3} - \sqrt{2})^2}{(\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2})}
Subtracting algebraic fractions requires finding the least common denominator, which is the product of the conjugate pair.
2
Expand the terms in the numerator
(\sqrt{3} + \sqrt{2})^2 = 3 + 2\sqrt{6} + 2 = 5 + 2\sqrt{6} \text{ and } (\sqrt{3} - \sqrt{2})^2 = 3 - 2\sqrt{6} + 2 = 5 - 2\sqrt{6}
Use the perfect square expansion formula (a±b)2=a2±2ab+b2(a \pm b)^2 = a^2 \pm 2ab + b^2.
3
Subtract the expanded terms in the numerator and simplify the denominator
\text{Numerator: } (5 + 2\sqrt{6}) - (5 - 2\sqrt{6}) = 4\sqrt{6}, \text{ Denominator: } (\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1
Apply the difference of two squares identity (ab)(a+b)=a2b2(a-b)(a+b) = a^2 - b^2 to the denominator and carefully distribute the negative sign across terms in the numerator.
4
Equate the simplified expression to k6k\sqrt{6} and solve for kk
k = 4
Comparing 461=46\frac{4\sqrt{6}}{1} = 4\sqrt{6} with k6k\sqrt{6} yields k=4k = 4.

Anahtar Kavram

Rationalisation of surd denominators using conjugate pairs and difference of squares
Soru 150Soru

The rate of heat transfer QQ across a building wall varies directly as the surface area AA of the wall and the temperature difference ΔT\Delta T between the interior and exterior, and inversely as the wall thickness dd. When the surface area is 4 m24\text{ m}^2, the temperature difference is 15C15^\circ\text{C}, and the thickness is 0.05 m0.05\text{ m}, the heat transfer rate is 1200 W1200\text{ W}. What is the heat transfer rate in watts when the surface area is 6 m26\text{ m}^2, the temperature difference is 20C20^\circ\text{C}, and the thickness is 0.08 m0.08\text{ m}?

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Cevap: 1500

Cevap

The heat transfer rate is 1500 W1500\text{ W}.
Establishing the variation constant k=1k = 1 using the initial given values and substituting the new parameters yields Q=1×6×200.08=1500 WQ = \frac{1 \times 6 \times 20}{0.08} = 1500\text{ W}.

Adım Adım Çözüm

1
Formulate the variation equation
Q=kAΔTdQ = \frac{k A \Delta T}{d}
Direct variation means multiplying by AA and ΔT\Delta T, while inverse variation means dividing by dd.
2
Calculate the constant of variation kk
k=1k = 1
Substituting Q=1200Q=1200, A=4A=4, ΔT=15\Delta T=15, and d=0.05d=0.05 gives 1200=60k0.05=1200k1200 = \frac{60k}{0.05} = 1200k, so k=1k = 1.
3
Compute the target heat transfer rate QQ
1500 W1500\text{ W}
Substituting k=1k=1, A=6A=6, ΔT=20\Delta T=20, and d=0.08d=0.08 gives Q=1×6×200.08=1200.08=1500 WQ = \frac{1 \times 6 \times 20}{0.08} = \frac{120}{0.08} = 1500\text{ W}.

Anahtar Kavram

Joint and Inverse Variation
Soru 151Soru

An arithmetic progression (A.P.) has a first term of 55 and a common difference of 3-3. What is the 7th7^{\text{th}} term of the progression?

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Cevap: 13-13

Cevap

13-13
Using the nthn^{\text{th}} term formula Tn=a+(n1)dT_n = a + (n - 1)d with a=5a = 5, d=3d = -3, and n=7n = 7, we get T7=5+(6)(3)=13T_7 = 5 + (6)(-3) = -13.

Adım Adım Çözüm

1
Identify the given parameters of the A.P.
First term a=5a = 5, common difference d=3d = -3, and position n=7n = 7.
These parameters are directly specified in the problem.
2
Apply the nthn^{\text{th}} term formula Tn=a+(n1)dT_n = a + (n - 1)d.
T7=5+(71)(3)=5+6(3)T_7 = 5 + (7 - 1)(-3) = 5 + 6(-3).
The common difference is added (n1)(n-1) times to the first term.
3
Perform the multiplication and addition.
T7=518=13T_7 = 5 - 18 = -13.
Simplifying the arithmetic gives the exact value of the 7th7^{\text{th}} term.

Anahtar Kavram

Calculating the nthn^{\text{th}} term of an Arithmetic Progression
Soru 152Soru

If (x1)(x - 1) is a factor of the polynomial P(x)=x3+2x25x+kP(x) = x^3 + 2x^2 - 5x + k, what is the value of kk?

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Cevap: 2

Cevap

The value of kk is 22.
According to the Factor Theorem, a linear polynomial (xa)(x - a) is a factor of P(x)P(x) if and only if P(a)=0P(a) = 0. For the factor (x1)(x - 1), setting x=1x = 1 gives P(1)=(1)3+2(1)25(1)+k=0P(1) = (1)^3 + 2(1)^2 - 5(1) + k = 0. Simplifying yields 1+25+k=01 + 2 - 5 + k = 0, which simplifies further to 2+k=0-2 + k = 0, giving k=2k = 2.

Adım Adım Çözüm

1
Apply the Factor Theorem
P(1)=0P(1) = 0
Since (x1)(x - 1) is a factor, setting x=1x = 1 makes the polynomial equal to zero.
2
Substitute x=1x = 1 into P(x)P(x)
13+2(1)25(1)+k=01^3 + 2(1)^2 - 5(1) + k = 0
Evaluate the polynomial expression at x=1x = 1.
3
Simplify and solve for kk
k=2k = 2
Combine constants 1+25=21 + 2 - 5 = -2 and solve 2+k=0-2 + k = 0.

Anahtar Kavram

Factor Theorem
Soru 153Soru

If 28103+7+43=k\sqrt{28 - 10\sqrt{3}} + \sqrt{7 + 4\sqrt{3}} = k, where kk is a rational number, find the value of kk.

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Cevap: 7

Cevap

The value of kk is 77.
Simplifying each nested surd into standard binomial surd form gives 28103=53\sqrt{28 - 10\sqrt{3}} = 5 - \sqrt{3} and 7+43=2+3\sqrt{7 + 4\sqrt{3}} = 2 + \sqrt{3}. Combining these terms cancels out the irrational component 3\sqrt{3}, leaving the integer result 5+2=75 + 2 = 7.

Adım Adım Çözüm

1
Simplify the nested surd 28103\sqrt{28 - 10\sqrt{3}} into binomial surd form
28103=53\sqrt{28 - 10\sqrt{3}} = 5 - \sqrt{3}
Assuming a form xy3x - y\sqrt{3} and squaring both sides gives x2+3y22xy3=28103x^2 + 3y^2 - 2xy\sqrt{3} = 28 - 10\sqrt{3}. Matching components leads to xy=5xy = 5 and x2+3y2=28x^2 + 3y^2 = 28, which yields integer values x=5x = 5 and y=1y = 1.
2
Simplify the nested surd 7+43\sqrt{7 + 4\sqrt{3}} into binomial surd form
7+43=2+3\sqrt{7 + 4\sqrt{3}} = 2 + \sqrt{3}
Assuming a form u+v3u + v\sqrt{3} and squaring both sides gives u2+3v2+2uv3=7+43u^2 + 3v^2 + 2uv\sqrt{3} = 7 + 4\sqrt{3}. Matching components leads to uv=2uv = 2 and u2+3v2=7u^2 + 3v^2 = 7, which yields integer values u=2u = 2 and v=1v = 1.
3
Sum the simplified expressions to calculate kk
k=7k = 7
Summing (53)+(2+3)(5 - \sqrt{3}) + (2 + \sqrt{3}) results in the irrational parts 3-\sqrt{3} and 3\sqrt{3} cancelling out, leaving 5+2=75 + 2 = 7.

Anahtar Kavram

Square Root of a Binomial Surd Expression
Soru 154Soru

The maximum safe load LL supported by a horizontal wooden beam varies directly as its width ww and the square of its depth dd, and inversely as its length ll. A beam of width 6 cm6\text{ cm}, depth 10 cm10\text{ cm}, and length 4 m4\text{ m} can support a maximum safe load of 900 kg900\text{ kg}. What is the maximum safe load that can be supported by a beam of the same material having a width of 4 cm4\text{ cm}, depth 12 cm12\text{ cm}, and length 6 m6\text{ m}?

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Cevap: 576 kg576\text{ kg}

Cevap

576 kg576\text{ kg}
The relationship is modeled by L=kwd2lL = \frac{k w d^2}{l}. Using the initial parameters (w=6w=6, d=10d=10, l=4l=4, L=900L=900), we find k=6k = 6. Substituting w=4w=4, d=12d=12, and l=6l=6 into the equation gives L=6×4×1446=576 kgL = \frac{6 \times 4 \times 144}{6} = 576\text{ kg}.

Adım Adım Çözüm

1
Formulate the variation equation
L=kwd2lL = \frac{k \cdot w \cdot d^2}{l}
Direct variation means multiplying factors in the numerator, while inverse variation places the variable in the denominator.
2
Calculate the constant of variation kk using initial conditions
900=k61024    900=600k4=150k    k=6900 = \frac{k \cdot 6 \cdot 10^2}{4} \implies 900 = \frac{600 k}{4} = 150 k \implies k = 6
Substitute L=900L = 900, w=6w = 6, d=10d = 10, and l=4l = 4 to solve for kk.
3
Calculate the new load LL for the new dimensions
L=641226=641446=576 kgL = \frac{6 \cdot 4 \cdot 12^2}{6} = \frac{6 \cdot 4 \cdot 144}{6} = 576\text{ kg}
Substitute k=6k = 6, w=4w = 4, d=12d = 12, and l=6l = 6 into the variation formula.

Anahtar Kavram

Joint and Inverse Variation
Tahmini Süre:1m 30s
Soru 155Soru

If (2x3)(2x - 3) is a factor of the polynomial P(x)=2x35x2+ax+6P(x) = 2x^3 - 5x^2 + ax + 6, what is the value of the constant aa?

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Cevap: -1

Cevap

The value of the constant aa is 1-1.
According to the Factor Theorem, if (2x3)(2x - 3) is a factor of P(x)P(x), then P(32)=0P\left(\frac{3}{2}\right) = 0. Substituting x=32x = \frac{3}{2} gives 274454+32a+6=0\frac{27}{4} - \frac{45}{4} + \frac{3}{2}a + 6 = 0, which simplifies to 32+32a=0\frac{3}{2} + \frac{3}{2}a = 0, resulting in a=1a = -1.

Adım Adım Çözüm

1
Apply the Factor Theorem by setting the linear factor equal to zero.
2x3=0    x=322x - 3 = 0 \implies x = \frac{3}{2}
If (2x3)(2x - 3) is a factor of P(x)P(x), then P(32)=0P\left(\frac{3}{2}\right) = 0.
2
Substitute x=32x = \frac{3}{2} into the polynomial expression P(x)P(x).
P(32)=2(32)35(32)2+a(32)+6=0P\left(\frac{3}{2}\right) = 2\left(\frac{3}{2}\right)^3 - 5\left(\frac{3}{2}\right)^2 + a\left(\frac{3}{2}\right) + 6 = 0
Setting the resulting expression equal to zero allows solving for aa.
3
Simplify the powers and numerical terms.
2(278)5(94)+32a+6=0    274454+32a+6=02\left(\frac{27}{8}\right) - 5\left(\frac{9}{4}\right) + \frac{3}{2}a + 6 = 0 \implies \frac{27}{4} - \frac{45}{4} + \frac{3}{2}a + 6 = 0
Evaluate each fraction before combining terms.
4
Combine constant terms and solve for aa.
184+6+32a=0    92+6+32a=0    32+32a=0    a=1-\frac{18}{4} + 6 + \frac{3}{2}a = 0 \implies -\frac{9}{2} + 6 + \frac{3}{2}a = 0 \implies \frac{3}{2} + \frac{3}{2}a = 0 \implies a = -1
Isolating aa yields the correct constant value.

Anahtar Kavram

Factor Theorem for linear divisors of the form (axb)(ax - b)
Soru 156Soru

What is the remainder when the polynomial P(x)=x34x2+5x2P(x) = x^3 - 4x^2 + 5x - 2 is divided by (x+1)(x + 1)?

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Cevap: 12-12

Cevap

12-12
According to the Remainder Theorem, dividing a polynomial P(x)P(x) by a linear divisor (xa)(x - a) yields a remainder equal to P(a)P(a). Here, the divisor is (x+1)(x + 1), which corresponds to a=1a = -1. Substituting x=1x = -1 into P(x)=x34x2+5x2P(x) = x^3 - 4x^2 + 5x - 2 gives (1)34(1)2+5(1)2=1452=12(-1)^3 - 4(-1)^2 + 5(-1) - 2 = -1 - 4 - 5 - 2 = -12. Thus, the value 12-12 is correct.

Adım Adım Çözüm

1
Apply the Remainder Theorem
To find the remainder when P(x)P(x) is divided by (x+1)(x + 1), evaluate P(1)P(-1) by setting x+1=0    x=1x + 1 = 0 \implies x = -1.
By the Remainder Theorem, dividing P(x)P(x) by (xa)(x - a) leaves a remainder equal to P(a)P(a).
2
Substitute x=1x = -1 into the polynomial P(x)=x34x2+5x2P(x) = x^3 - 4x^2 + 5x - 2
P(1)=(1)34(1)2+5(1)2P(-1) = (-1)^3 - 4(-1)^2 + 5(-1) - 2
Replace each instance of xx with 1-1.
3
Simplify the powers and terms
P(1)=14(1)52=1452=12P(-1) = -1 - 4(1) - 5 - 2 = -1 - 4 - 5 - 2 = -12
Compute arithmetic operations following standard order of operations.

Anahtar Kavram

Remainder Theorem
Tahmini Süre:45s
Soru 157Soru

If yy varies inversely as xx, and y=8y = 8 when x=3x = 3, what is the value of yy when x=6x = 6?

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Cevap: 4

Cevap

4
For inverse variation, y=kxy = \frac{k}{x}. Given y=8y = 8 when x=3x = 3, the constant of variation k=8×3=24k = 8 \times 3 = 24. Substituting x=6x = 6 gives y=246=4y = \frac{24}{6} = 4.

Adım Adım Çözüm

1
Set up the equation for inverse variation
y=kxy = \frac{k}{x}
Inverse variation means yy is inversely proportional to xx with constant kk.
2
Solve for the variation constant kk using y=8y = 8 and x=3x = 3
k=8×3=24k = 8 \times 3 = 24
Multiply both sides of the equation by xx.
3
Calculate yy when x=6x = 6
y=246=4y = \frac{24}{6} = 4
Substitute the constant k=24k = 24 and x=6x = 6 into the inverse variation equation.

Anahtar Kavram

Inverse Variation
Soru 158Soru

The electric power PP dissipated in a resistor varies directly as the square of the current II flowing through it. If a current of 3 A3\text{ A} produces a power of 45 W45\text{ W}, what is the power dissipated, in watts, when the current is 5 A5\text{ A}?

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Cevap: 125

Cevap

The power dissipated when the current is 5 A is 125 W.
Because electric power varies directly as the square of the current, the formula is P=kI2P = k I^2. Substituting the initial conditions gives 45=k(32)=9k45 = k(3^2) = 9k, so k=5k = 5. Evaluating at I=5 AI = 5\text{ A} gives P=5(52)=125 WP = 5(5^2) = 125\text{ W}.

Adım Adım Çözüm

1
Set up the variation equation
P=kI2P = k I^2
Power varies directly as the square of current.
2
Find the constant of variation kk
k=5k = 5
Substitute P=45P = 45 and I=3I = 3 into the variation equation: 45=k(32)    45=9k    k=545 = k(3^2) \implies 45 = 9k \implies k = 5.
3
Calculate the required power for I=5 AI = 5\text{ A}
P=125 WP = 125\text{ W}
Substitute k=5k = 5 and I=5I = 5 into P=kI2P = k I^2: P=5(52)=125P = 5(5^2) = 125.

Anahtar Kavram

Direct variation involving a squared quantity
Soru 159Soru

The polynomial P(x)=2x3+ax2+bx+6P(x) = 2x^3 + ax^2 + bx + 6 leaves a remainder of 1212 when divided by (x1)(x - 1) and has (x+3)(x + 3) as a factor. What is the value of aba - b?

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Cevap: 66

Cevap

The value of aba - b is 66.
Applying the Remainder Theorem with P(1)=12P(1) = 12 gives a+b=4a + b = 4. Applying the Factor Theorem with P(3)=0P(-3) = 0 gives 3ab=163a - b = 16. Solving these simultaneous linear equations gives a=5a = 5 and b=1b = -1. Evaluating ab=5(1)a - b = 5 - (-1) gives 66.

Adım Adım Çözüm

1
Apply the Remainder Theorem for the divisor (x1)(x - 1)
a+b=4a + b = 4
According to the Remainder Theorem, P(1)=12P(1) = 12. Substituting x=1x = 1 gives 2(1)3+a(1)2+b(1)+6=12    a+b+8=12    a+b=42(1)^3 + a(1)^2 + b(1) + 6 = 12 \implies a + b + 8 = 12 \implies a + b = 4.
2
Apply the Factor Theorem for the factor (x+3)(x + 3)
3ab=163a - b = 16
According to the Factor Theorem, P(3)=0P(-3) = 0. Substituting x=3x = -3 gives 2(3)3+a(3)2+b(3)+6=0    54+9a3b+6=0    9a3b=48    3ab=162(-3)^3 + a(-3)^2 + b(-3) + 6 = 0 \implies -54 + 9a - 3b + 6 = 0 \implies 9a - 3b = 48 \implies 3a - b = 16.
3
Solve the system of linear equations for aa and bb
a=5a = 5 and b=1b = -1
Adding (a+b=4)(a + b = 4) and (3ab=16)(3a - b = 16) yields 4a=20    a=54a = 20 \implies a = 5. Substituting a=5a = 5 into a+b=4a + b = 4 gives b=1b = -1.
4
Calculate the target value aba - b
ab=6a - b = 6
Subtracting bb from aa gives 5(1)=5+1=65 - (-1) = 5 + 1 = 6.

Anahtar Kavram

Polynomial Remainder and Factor Theorems
Soru 160Soru

In a survey of 160 music enthusiasts, 80 listen to Afrobeat, 70 listen to Highlife, and 65 listen to Reggae. It was found that 30 listen to both Afrobeat and Highlife, 25 listen to both Highlife and Reggae, and 28 listen to both Afrobeat and Reggae. If 15 enthusiasts listen to none of these three genres, how many enthusiasts listen to exactly two of these genres?

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Cevap: 44

Cevap

The number of enthusiasts who listen to exactly two of the three genres is 44.
The correct answer is 44. By subtracting the 15 non-listeners from 160, the union of the three sets contains 145 members. Solving the 3-set inclusion-exclusion equation yields 13 enthusiasts who listen to all three genres. Subtracting 13 from each given pairwise intersection gives 17, 12, and 15 for the regions representing listeners of exactly two genres. Summing these three counts yields 44.

Adım Adım Çözüm

1
Calculate the cardinality of the union of all three sets.
n(AHR)=16015=145n(A \cup H \cup R) = 160 - 15 = 145
Subtracting the number of enthusiasts listening to none of the genres from the universal set.
2
Apply the Principle of Inclusion-Exclusion for 3 sets to find the triple intersection.
n(AHR)=13n(A \cap H \cap R) = 13
Substituting known cardinalities into n(AHR)=n(A)+n(H)+n(R)[n(AH)+n(HR)+n(AR)]+n(AHR)n(A \cup H \cup R) = n(A)+n(H)+n(R) - [n(A \cap H)+n(H \cap R)+n(A \cap R)] + n(A \cap H \cap R).
3
Calculate the count of enthusiasts in each 'exactly two genres' region.
Afrobeat and Highlife only = 17; Highlife and Reggae only = 12; Afrobeat and Reggae only = 15.
Subtracting the triple intersection count (1313) from each pairwise intersection.
4
Sum the counts of the three distinct two-genre regions.
17+12+15=4417 + 12 + 15 = 44
Combining all mutually exclusive regions representing enthusiasts of exactly two genres.

Anahtar Kavram

Principle of Inclusion-Exclusion for three sets and Venn diagram region partitioning.
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Algebra Alıştırma Soruları — JAMB UTME — Sayfa 8 | Examkin