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2 years ago

How do you solve this equation: 80x =

Mathematics

1 answer:

Pepsi [2]2 years ago

7 0

Answer:

Divide by 80 on both sides

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A building is 50 feet tall. Imagine you are standing 30 feet away from it. What is the angle when you are looking up at your fri

Rina8888 [55]

Answer:

53.1

Step-by-step explanation:

I used my protractor and that is what I got there is no detailed explanation that I came up with

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3 years ago

Somebody please help asapppppp

anzhelika [568]

The function was moved two units to the left (a)

8 0

3 years ago

B) Find g (-3) for g(x) = 3x + 5

fredd [130]

Step-by-step explanation:

g(x)=3x+5

g(-3)=3(-3)+5

=-9+5

=-4

3 0

2 years ago

149 cars are waiting to take a ferry across the channel. Each ferry can only hold 18 cars. How many cars will be on the last fer

balandron [24]

$149 \div 18 = x$
$8r5 = x$
There will be 9 trips the ferry makes and on the last trip there will be 5 cars on the ferry.

5 0

3 years ago

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