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3 years ago

NEED HELP ASAP!!! 5-10 points Number three, what is the answer?

Mathematics

1 answer:

liraira [26]3 years ago

6 0

Dario went -60 feet below

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Please help! URGENT! will award 30 points & brainliest!

Alekssandra [29.7K]

Answer:

i believe it is -21, but i'm not sure

Step-by-step explanation:

just foil the two to get a quadratic expression

4 0

3 years ago

(9x-9)(9×+9), your finding the difference between two squares. im have some trouble in how to solve it.

sergejj [24]

Answer: Your answer is 0 (F.O.I.L) first. outer. inner. last. Or you could use what my teacher calls the magic box

Step-by-step explanation:

So for the boxes you multiply for example the top left box you multiply the 9 above it and the 9 to the left or like the top right box you multiply the -9 above it and the 9 on the outside of the box to the left

8 0

3 years ago

What happens if you label the starting number x

tester [92]

You can use any letter but when you label it x you find x

3 0

3 years ago

Can anyone help me on my HW i can't seem to find the answer to it and when i do it doesn't make sence

Nadusha1986 [10]

0.75x-18.5=0.65x
0.75x-0.65x=18.5
0.1x=18.5
x=18.5/0.1=185

Answer: x=185

To checkt:
0.75(185)-18.5=138.75-18.5=120.25
0.65(185)=120.25

4 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

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