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Lina20 [59]
4 years ago
11

Which of the following is the most appropriate unit to describe the time you can drive a car based on the amount of gas?

Mathematics
2 answers:
Anton [14]4 years ago
7 0
I believe the answer to the question is A. This is because the minutes rely on the amount of gas that you have.
3241004551 [841]4 years ago
6 0
Yo i think it's A, due to the fact that the more gallons you have, the more minutes you can drive. hope this helps, have an amazing day :)
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This rectangle is half as wide as it is long what is the perimeter of the rectangle
kramer

length=4 cm width=2 cm

4(2)+2(2)=12

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3 years ago
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34. The product of two algebraic
S_A_V [24]

Answer:

3a²b

Step-by-step explanation:

The product of two algebraic

terms is 6a3b2. If one of the terms is

2ab, find the other term.

Let us represent

First term = a

Other term = b

a × b = 6a³b²

a = 2ab

b = ?

b = 6a³b²/a

b = 6a³b²/2ab

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Therefore, the other term = 3a²b

7 0
3 years ago
What is x if 4x-79=3x-29
elena-14-01-66 [18.8K]
<em>4x-79=3x-29</em>

<em>Move the terms with "x" on the left, and the numbers on the right.</em>

<em>4x-3x=79-29</em>
<em>x=50</em>
6 0
3 years ago
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If <img src="https://tex.z-dn.net/?f=%20300cm%5E%7B2%7D%20" id="TexFormula1" title=" 300cm^{2} " alt=" 300cm^{2} " align="absmid
Artist 52 [7]
Check the picture below.  Recall, is an open-top box, so, the top is not part of the surface area, of the 300 cm².  Also, recall, the base is a square, thus, length = width = x.

\bf \textit{volume of a rectangular prism}\\\\&#10;V=lwh\quad &#10;\begin{cases}&#10;l = length\\&#10;w=width\\&#10;h=height\\&#10;-----\\&#10;w=l=x&#10;\end{cases}\implies V=xxh\implies \boxed{V=x^2h}\\\\&#10;-------------------------------\\\\&#10;\textit{surface area}\\\\&#10;S=4xh+x^2\implies 300=4xh+x^2\implies \cfrac{300-x^2}{4x}=h&#10;\\\\\\&#10;\boxed{\cfrac{75}{x}-\cfrac{x}{4}=h}\\\\&#10;-------------------------------\\\\&#10;V=x^2\left( \cfrac{75}{x}-\cfrac{x}{4} \right)\implies V(x)=75x-\cfrac{1}{4}x^3

so.. that'd be the V(x) for such box, now, where is the maximum point at?

\bf V(x)=75x-\cfrac{1}{4}x^3\implies \cfrac{dV}{dx}=75-\cfrac{3}{4}x^2\implies 0=75-\cfrac{3}{4}x^2&#10;\\\\\\&#10;\cfrac{3}{4}x^2=75\implies 3x^2=300\implies x^2=\cfrac{300}{3}\implies x^2=100&#10;\\\\\\&#10;x=\pm10\impliedby \textit{is a length unit, so we can dismiss -10}\qquad \boxed{x=10}

now, let's check if it's a maximum point at 10, by doing a first-derivative test on it.  Check the second picture below.

so, the volume will then be at   \bf V(10)=75(10)-\cfrac{1}{4}(10)^3\implies V(10)=500 \ cm^3

6 0
3 years ago
What is the distance between point A and point B to the nearest tenth?
Lyrx [107]
What is the number line i'm sorry but i can help without a number line that this question is based off of.
5 0
3 years ago
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