All categories

3 years ago

Pls help me guys! This is soooo hard!

Mathematics

1 answer:

aliya0001 [1]3 years ago

7 0

B, because there are no 30-39 and two 20-29

You might be interested in

Please please please help!!!

dalvyx [7]

9514 1404 393

Answer:

Step-by-step explanation:

Fill in the argument value and look up the function value on the graph.

$(f\circ f)(-2) = f(f(-2)) = f(2) = \boxed{2}$

7 0

2 years ago

I bought a melon that weighed 1 whole 3/5 lb I cut a piece that weighed 4/5 pounds and gave it to my neighbor I had one fifth po

user100 [1]

Answer: 3/5

Step-by-step explanation:

if you start off with 5/5 + 3/5, thats 8/5 - 4/5 is 4/5, - 1/5 is 3/5

6 0

3 years ago

Please show me how to work this out

4vir4ik [10]

Answer:

f(x) = -4x + 12

Step-by-step explanation:

y - 2x = -6x + 12

+2x +2x

f(x) = -4x + 12

3 0

3 years ago

A right triangle has a leg b of length 7 and a hypotenuse of length 11 what is the length of the other leg a?

scoray [572]

Answer: 8.5

Step-by-step explanation:

5 0

2 years ago

Sand pouring from a chute forms a conical pile whose height is always equal to the diameter. If the height increases at a consta

Cerrena [4.2K]

Answer:

$125\pi \frac{\text{ft}^3}{\text{min}}$ .

Step-by-step explanation:

Let x represent height of the cone.

We have been given that Sand pouring from a chute forms a conical pile whose height is always equal to the diameter.

We know that radius is half the diameter, so radius of cone would be $\frac{x}{2}$ .

We will use volume of cone formula to solve our given problem.

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

Upon substituting the value of height and radius in terms of x, we will get:

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

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

$V=\frac{1}{12}\pi x^3$

Now, we will take the derivative of volume with respect to time as:

$\frac{dV}{dt}=\frac{1}{12}\pi\cdot 3x^2\cdot \frac{dx}{dt}$

$\frac{dV}{dt}=\frac{1}{4}\pi\cdot x^2\cdot \frac{dx}{dt}$

Upon substituting $x=10$ and $\frac{dx}{dt}=5$ , we will get:

$\frac{dV}{dt}=\frac{1}{4}\pi\cdot (10)^2\cdot 5$

$\frac{dV}{dt}=\frac{1}{4}\pi\cdot 100\cdot 5$

$\frac{dV}{dt}=\pi\cdot 25\cdot 5$

$\frac{dV}{dt}=125\pi$

Therefore, the sand is pouring from the chute at a rate of $125\pi \frac{\text{ft}^3}{\text{min}}$ .

8 0

3 years ago