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2 years ago

How long should an escalator be if it is to make an angle 30 with the floor and

Mathematics

1 answer:

Charra [1.4K]2 years ago

7 0

Answer:

it should be 23 feet long

Step-by-step explanation:

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I need help remembering the steps for 2x+5y=-7

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Get Y by itself so it' should look like this

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Y=2/5x+7/5

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3 years ago

In Exercises 9 and 10, point B is between A and Con AC. Use the information to

Marizza181 [45]

Answer:n

Step-by-step explanation:

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3 years ago

A PACK OF TOILET PAPER WAS ORIGINALLY $4.50 , IT COST $6.75. DID THE PRICE INCREASE OR DECREASE?

Lady bird [3.3K]

Increase. If the toilet paper is originally 4.50 and it then is 6.50 there is a price difference. It increased.

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2 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

3 years ago

Find the standard form of the equation of the hyperbola satisfying the given conditions: X intercept +/- 6; foci at (-10,0) and

uysha [10]

Answer:

$\frac{x^{2}}{36} - \frac{y^{2}}{64}=1$

Step-by-step explanation:

Given an hyperbola with the following conditions:

Foci at (-10,0) and (10,0)
x-intercept +/- 6;

The following holds:

The center is midway between the foci, so the center must be at (h, k) = (0, 0).

The foci are 10 units to either side of the center, so c = 10 and $c^2 = 100$
The center lies on the origin, so the two x-intercepts must then also be the hyperbola's vertices.

Since the intercepts are 6 units to either side of the center, then a = 6 and $a^2 = 36.$

$Then, a^2+b^2=c^2\\b^2=100-36=64$

Therefore, substituting $a^2 = 36.$ and $b^2=64$ into the standard form

$\frac{x^{2}}{a^2} - \frac{y^{2}}{b^2}=1\\We \: have:\\ \dfrac{x^{2}}{36} - \dfrac{y^{2}}{64}=1$

4 0

3 years ago