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vampirchik [111]
2 years ago
14

52 in 39 in What is the length of the hypotenuse? C = inches

Mathematics
1 answer:
Lesechka [4]2 years ago
6 0
C, the hypotenuse is 65 in
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Simplify the expression.<br> 7h+(8.9d)-17+6d-3.4h
zhuklara [117]

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Step-by-step explanation:

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2 years ago
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a group of 266 people is called to jury duty in court. Each jury includes 12 jurors plus to alternates. How many complete juries
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There are 14 positions. There are 266 choices for the first juror, 265 for the second, 264 for the third, etc. 266*265*264*...*252=<span>5.93893009829e+33, or about 6,000,000,000,000,000,000,000,000,000,000,000 complete juries. Hope this helps!</span>
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3 years ago
Chris and Becky are comparing mortgages. The mortgage is for $183,500. If they choose 30 years at 5%, their monthly payment will
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For 30 years at 5% = $354,625.2

For 20 years at 4.5% = $278,618.4

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5 0
3 years ago
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What is the solution of 6(3-2x)=54?
notka56 [123]
Hello!

You solve this algebraically

6(3 - 2x) = 54

distribute the 6

18 - 12x = 54

Subtract 18 from both sides

-12x = 36

Divide both sides by -12

x = -3

The answer is -3

Hope this helps!
3 0
3 years ago
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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
3 years ago
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