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3 years ago

10

8cups for $4 how much for 1 cup

1 answer:

Ilya [14]3 years ago

4 0

0.5

4/8=0.5

Check work: 0.5•8=$4

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the volume of the tent is 30ft³

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An inverted pyramid is being filled with water at a constant rate of 25 cubic centimeters per second. The pyramid, at the top, h

UNO [17]

Answer:

$\frac{225}{16} cm/s$

Step-by-step explanation:

We are given that

$\frac{dV}{dt}=25cm^3/s$

Side of base=4 cm

l=w=4 cm

Height,h=12 cm

We have to find the rate at which the water level rising when the water level is 4 cm.

Volume of pyramid= $\frac{1}{3}lwh=\frac{1}{3}l^2h$

$\frac{l}{h}=\frac{4}{12}=\frac{1}{3}$

$l=\frac{1}{3}h$

Substitute the value

$V=\frac{1}{27}h^3$

Differentiate w.r.t t

$\frac{dV}{dt}=\frac{3}{27}h^2\frac{dh}{dt}$

Substitute the values

$25=\frac{1}{9}(4^2)\frac{dh}{dt}$

$\frac{dh}{dt}=\frac{25\times 9}{16}=\frac{225}{16} cm/s$

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3 years ago

At 3pm, the length of the shadow of a thin vertical pole standing on level ground is the same as the height of the pole. A while

aliya0001 [1]

Answer: The height of the pole is 175.97 ( approx )

Step-by-step explanation:

Let the height of the pole is x cm,

Also, let the angle of elevation of the sun to the pole at 3 pm is $\theta$

Thus, by the question,

$tan\theta = \frac{\text{ The height of the pole}}{\text{ The height of the shadow}}$

$\implies tan\theta = \frac{x}{x}$ ( at 3pm the height of shadow = height of the pole)

$\implies tan\theta = 1$

$\implies \theta = 45^{\circ}$

Again, according to the question,

When the angle of elevation is $(\theta - 12^{\circ})$ ,

The height of shadow = x + 95,

$\implies tan(45-12)^{\circ}=\frac{x}{x+95}$

$\implies tan 33^{\circ}=\frac{x}{x+95}$

$\implies tan 33^{\circ}x + 95\times tan33^{\circ}=x$

$\implies tan 33^{\circ}x - x = - 95\times tan33^{\circ}$

$\implies -0.3505924068 x = - 61.6937213538$

$x=175.969930202\approx 175.97\text{ cm}$

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3 years ago

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If there isn't a base, the base is 10 :)

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