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3 years ago

Charles can move 16 boxes in 4/3 of an hour. How many boxes can he move in one hour

Mathematics

2 answers:

r-ruslan [8.4K]3 years ago

5 0

Answer:

Your answer is 4

Step-by-step explanation:

In one hour he made 4 boxes two hours he made 8 just add four from there

kondaur [170]3 years ago

4 0

Answer:

12 boxes in 1 hour

Step-by-step explanation:

he moves 16 boxes in 1 and a third hour.

1/3 hour = 4 boxes so one hour would be 12 boxes.

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Answer:

it going down dividing by 2

Step-by-step explanation:

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3 years ago

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Otrada [13]

Answer:

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Step-by-step explanation:

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3 years ago

Write the equation of the line<br> Screenshot attached

andrew-mc [135]

Answer:

y = -3/2x + 4

Step-by-step explanation:

Do rise over run to get the -3/2x

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6 0

3 years ago

Read 2 more answers

Statements

vazorg [7]

Answer:

See Explanation

Step-by-step explanation:

The question has unclear information.

So, I'll answer from scratch

Given

ABC = Right angled triangle

DB bisects ABC

Required

Prove that CBD = 45

From the question, we have that:

ABC is right angled at B

So, when DB bisects ABC, it means that DB divides ABC into two equal angles.

i.e.

$CBD = ABD$

and

$CBD + ABD = 90$

Substitute CBD for ABD in $CBD + ABD = 90$

$CBD + CBD = 90$

$2CBD = 90$

Divide both sides by 2

$\frac{2CBD}{2} = \frac{90}{2}$

$CBD = \frac{90}{2}$

$CBD = 45$

Hence, it is proved that $CBD = 45$

<em>Follow the above explanation and use it to answer your question properly</em>

6 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