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Fittoniya [83]
3 years ago
13

What is a equivalent expressions 3(6x+4)

Mathematics
1 answer:
Contact [7]3 years ago
4 0

Answer: 18x + 12

Step-by-step explanation:

3 ( 6x + 4 )

3 x 6x = 18x

3 x 4 = 12

18x + 12

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Suppose that a student needs to buy 6 books for her history course. the number of books that she will be able to find used is a
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Probability is approximately: 0.256



8 0
3 years ago
Check out the photo for the question
Anestetic [448]

Answer:

50

Step-by-step explanation:

42/n³∑k²+12/n²∑k+30/n∑1

=42/n³[n(n+1)(2n+1)/6]+12/n²[n(n+1)/2]+30/n [n]

=7n(n+1)(2n+1)/n³+6n(n+1)/n²+30

=7(n+1)(2n+1)/n²+6(n+1)/n+30

=[7(2n²+3n+1)+6(n²+n)+30n²]/n²

=[14n²+21n+7+6n²+6n+30n²]/n²

=[50n²+27n+7]/n²

=[50+27/n+7/n²]

→50 as n→∞

because 1/n,1/n²→0 as n→∞

4 0
4 years ago
Read 2 more answers
Subtract. Use models if needed.<br><br><br> (12x + 3) – (2x + 10)
Galina-37 [17]
1.)combine like terms: 12-2x=10x 3-10=-7

10x-7
4 0
3 years ago
Read 2 more answers
Remember to show work and explain. Use the math font.
MrMuchimi

Answer:

\large\boxed{1.\ f^{-1}(x)=4\log(x\sqrt[4]2)}\\\\\boxed{2.\ f^{-1}(x)=\log(x^5+5)}\\\\\boxed{3.\ f^{-1}(x)=\sqrt{4^{x-1}}}

Step-by-step explanation:

\log_ab=c\iff a^c=b\\\\n\log_ab=\log_ab^n\\\\a^{\log_ab}=b\\\\\log_aa^n=n\\\\\log_{10}a=\log a\\=============================

1.\\y=\left(\dfrac{5^x}{2}\right)^\frac{1}{4}\\\\\text{Exchange x and y. Solve for y:}\\\\\left(\dfrac{5^y}{2}\right)^\frac{1}{4}=x\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\\dfrac{(5^y)^\frac{1}{4}}{2^\frac{1}{4}}=x\qquad\text{multiply both sides by }\ 2^\frac{1}{4}\\\\\left(5^y\right)^\frac{1}{4}=2^\frac{1}{4}x\qquad\text{use}\ (a^n)^m=a^{nm}\\\\5^{\frac{1}{4}y}=2^\frac{1}{4}x\qquad\log_5\ \text{of both sides}

\log_55^{\frac{1}{4}y}=\log_5\left(2^\frac{1}{4}x\right)\qquad\text{use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\\dfrac{1}{4}y=\log(x\sqrt[4]2)\qquad\text{multiply both sides by 4}\\\\y=4\log(x\sqrt[4]2)

--------------------------\\2.\\y=(10^x-5)^\frac{1}{5}\\\\\text{Exchange x and y. Solve for y:}\\\\(10^y-5)^\frac{1}{5}=x\qquad\text{5 power of both sides}\\\\\bigg[(10^y-5)^\frac{1}{5}\bigg]^5=x^5\qquad\text{use}\ (a^n)^m=a^{nm}\\\\(10^y-5)^{\frac{1}{5}\cdot5}=x^5\\\\10^y-5=x^5\qquad\text{add 5 to both sides}\\\\10^y=x^5+5\qquad\log\ \text{of both sides}\\\\\log10^y=\log(x^5+5)\Rightarrow y=\log(x^5+5)

--------------------------\\3.\\y=\log_4(4x^2)\\\\\text{Exchange x and y. Solve for y:}\\\\\log_4(4y^2)=x\Rightarrow4^{\log_4(4y^2)}=4^x\\\\4y^2=4^x\qquad\text{divide both sides by 4}\\\\y^2=\dfrac{4^x}{4}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\y^2=4^{x-1}\Rightarrow y=\sqrt{4^{x-1}}

6 0
4 years ago
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8_murik_8 [283]

Answer:

d

Step-by-step explanation:

5 0
3 years ago
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