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Ainat [17]
2 years ago
13

The volume of a rectangular box is 1600 cubic inches. The height of the box is 8 inches and the length of the box is 2 times lon

ger than the width of the box. What is the width of the box "IN FEET FT"?
Mathematics
1 answer:
Anastasy [175]2 years ago
7 0

Answer:

0.833 feets

Step-by-step explanation:

Given that:

Volume of box = 1600

Heigh of box (H) = 8 inches

Let:

Width of box (W) = x

Length of box (L) = 2x

Volume of box = (Height * width * length)

1600 = (8 * x * 2x)

1600 = 16x²

x² = 1600/16

x² = 100

x = sqrt(100)

x = 10 inches

Hence, width of box in feet:

1 inch = 0.0833 feets

10 inches = (10 * 0.0833) feets

= 0.833 feets

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Xiao bought ​ 4310 ​ gallons of gas that cost $2.50 per gallon.
Vikentia [17]

Answer:

$10775

Step-by-step explanation:

If 1 gallon of gas is $2.50, then the cost of 4310 gallons of gas will be:

4310 * 2.50

= $10775

Hope this answer was helpful!!

8 0
3 years ago
Read 2 more answers
Please help me with this geometry question:((
Len [333]

Answer:

x = 13

Arc RQ = 180°

Step-by-step explanation:

PEQ is an inscribed angle and the measure of an inscribed angle in a circle is equal to half of the arc it sees.

The perimeter of a circle is always 360° therefore

2 × (5x + 15) + 9x + 17 + 6x - 12 = 25x + 35 = 360°

25x = 360 - 35

25x = 325

x = 13

Since arc RQ is equal to 10x + 30 we replace x with the value we just got

10×13 + 30 = 160°

8 0
2 years ago
Read 2 more answers
Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"
vampirchik [111]
\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h

Employ a standard trick used in proving the chain rule:

\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h

The limit of a product is the product of limits, i.e. we can write

\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)

The rightmost limit is an exercise in differentiating \sqrt x using the definition, which you probably already know is \dfrac1{2\sqrt x}.

For the leftmost limit, we make a substitution y=\sqrt x. Now, if we make a slight change to x by adding a small number h, this propagates a similar small change in y that we'll call h', so that we can set y+h'=\sqrt{x+h}. Then as h\to0, we see that it's also the case that h'\to0 (since we fix y=\sqrt x). So we can write the remaining limit as

\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}

which in turn is the derivative of \tan y, another limit you probably already know how to compute. We'd end up with \sec^2y, or \sec^2\sqrt x.

So we find that

\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}
7 0
3 years ago
Andy is blowing up a spherically shaped balloon. If he is able to blow 81π cm^3 of air with every breath, it takes him 12 breath
Allisa [31]

The radius of the balloon given the information in the question is 8.98cm.

<h3>What is the radius of the balloon?</h3>

The first step is to determine the volum of the balloon.

Volume = 81π cm^3 x 12 = 972π cm^3

Now, determine the radius using this formula:

∛[Volume / (4/3π )]

∛[972π cm^3/ (4/3π )] = 8.98cm

Here is the complete question:

Andy is blowing up a spherically shaped balloon. If he is able to blow 81π cm^3 of air with every breath, it takes him 12 breaths to fully inflate the balloon. What is the radius of the balloon?

To learn more about the volume of a sphere, please check: brainly.com/question/13705125

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1 year ago
Evaluate -(a + b), given values a = -4, b = - 12.
tatuchka [14]
The answer would be -16
4 0
3 years ago
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