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3 years ago

11

Find the surface area of a box measuring 4 inches long 3 inches wide and 6 inches tall

1 answer:

Rufina [12.5K]3 years ago

3 0

4*3

12

12*6

72

hope it help

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3 years ago

Graph the line with slope <br> 2<br> 5<br><br> and <br> y<br><br> -intercept <br> −<br> 7<br><br> .

tia_tia [17]

Answer:

See attachment

Step-by-step explanation:

We want to graph the line with slope

$m = \frac{2}{5}$

and y-intercept ,

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$y = \frac{2}{5}x - 7$

Starting at -7 on the y-axis, we run 5 units and rise 2 units right.

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The graph is shown in the attachment.

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3 years ago

Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 4cm and a height of 16cm, at the ra

bearhunter [10]

Answer:

$\frac{dh}{dt}=-\frac{1}{2\pi}cm/min$

Step-by-step explanation:

From similar triangles, see diagram in attachment

$\frac{r}{4}=\frac{h}{16}$

We solve for $r$ to obtain,

$r=\frac{h}{4}$

The formula for calculating the volume of a cone is

$V=\frac{1}{3}\pi r^2h$

We substitute the value of $r=\frac{h}{4}$ to obtain,

$V=\frac{1}{3}\pi (\frac{h}{4})^2h$

This implies that,

$V=\frac{1}{48}\pi h^3$

We now differentiate both sides with respect to $t$ to get,

$\frac{dV}{dt}=\frac{\pi}{16}h^2 \frac{dh}{dt}$

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This implies that $\frac{dV}{dt}=-2cm^3/min$ .

Since we want to determine the rate at which the depth of the water is changing at the instance when the water in the tank is 8cm deep, it means $h=8cm$ .

We substitute this values to obtain,

$-2=\frac{\pi}{16}(8)^2 \frac{dh}{dt}$

$\Rightarrow -2=4\pi \frac{dh}{dt}$

$\Rightarrow -1=2\pi \frac{dh}{dt}$

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3 years ago

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