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2 years ago

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Last season, a volleyball team won 40\%40% of 8080 games played.

1 answer:

IRISSAK [1]2 years ago

8 0

Answer:

Happy Chrismas friend!

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Find the quotient of 1 1/4 and 3 1/2 . Express your answer in simplest form.

tino4ka555 [31]

Answer:

8/35

Step-by-step explanation:

1 1/4 = 5/4

3 1/2 = 7/2

5/4 × 7/2

Find the reciprocal. (flip numerator and denomimator)

4/5 × 2/7 = 8/35

4 0

2 years ago

Adele is paid a weekly salary of $685. She is paid an additional $23.50 for every hour of overtime she works. This week her tota

loris [4]

The answer is 7 overtime hours

7 0

2 years ago

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What is the difference of polynomials (8r6s3-9r5s4+3r4s5)-(2r4s-5r6-4r5s4)

Alinara [238K]

If you would like to solve <span>(8r^6s^3 – 9r^5s^4 + 3r^4s^5) – (2r^4s^5 – 5r^3s^6 – 4r^5s^4), you can do this using the following steps:

</span>(8r^6s^3 – 9r^5s^4 + 3r^4s^5) – (2r^4s^5 – 5r^3s^6 – 4r^5s^4) = 8r^6s^3 – 9r^5s^4 + 3r^4s^5 – 2r^4s^5 + 5r^3s^6 + 4r^5s^4 = 8r^6s^3 – 5r^5s^4 + r^4s^5<span> + 5r^3s^6
</span>
The correct result would be 8r^6s^3 – 5r^5s^4 + r^4s^5<span> + 5r^3s^6.</span>

7 0

2 years ago

I'm stuck on this question! Please help

gayaneshka [121]

See attachment file below.
Answer: 11/3

Hope it helped!

6 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

3 years ago

Read 2 more answers