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4 years ago

6

Find the equation of quadratic function with (6,-2) and (5,-5)

1 answer:

aliina [53]4 years ago

3 0

Answer:

y= 3(x-5)^2-5 is the equation.

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3 years ago

Simplify -2a(7a^5-5a+5)

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2 years ago

Is the following relation a function? <br><br> A. Yes<br> B. No

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3 years ago

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Anita has 2 quarters, 1 nickel, and 4 pennies. Write Anita’s total amount as a fraction of a dollar.

KatRina [158]

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5 0

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,

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

We were given that water is drained out of the tank at a rate of $2cm^3/min$

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,

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

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

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

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

3 0

4 years ago

Read 2 more answers