Ask question

Ask question

All categories

3 years ago

10

Please help me solve 9 and 10 ASAP please

1 answer:

kipiarov [429]3 years ago

8 0

Answer:

$\sqrt{ |?| }$

You might be interested in

Sean bought a 6 inch sub. He ate 3 3/4 inches. How much sub does he have left in the simplest form.

irinina [24]

Answer: 2 1/2

hope this helps

4 0

3 years ago

A spherical balloon is being inflated at a rate of 3 cubic inches per second. Determine the change in the rate of the radius.How

Novay_Z [31]

Answer:

The rate rate of change of radius is $\frac{1}{48\pi}$ inches per second when the diameter is 12 inches.

The radius is changing more rapidly when the diameter is 12 inches.

Step-by-step explanation:

Consider the provided information.

A spherical balloon is being inflated at a rate of 3 cubic inches per second.

The volume of sphere is $V=\frac{4}{3}\pi r^3$

Differentiate the above formula with respect to time.

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$

Substitute the respective values in the above formula,

$3=4\pi 6^2\frac{dr}{dt}$

$\frac{1}{48\pi}=\frac{dr}{dt}$

The rate rate of change of radius is $\frac{1}{48\pi}$ inches per second when the diameter is 12 inches.

When d=16

$3=4\pi 8^2\frac{dr}{dt}$

$\frac{3}{256\pi}=\frac{dr}{dt}$

Thus, the radius is changing more rapidly when the diameter is 12 inches.

5 0

3 years ago

I need the angle of elevation

Aleksandr-060686 [28]

Answer:

multiply and see what you get

4 0

3 years ago

In triangle PQR, PR = 23mm, QR = 39mm, and m<R = 163 degrees. Find the area of the triangle to the nearest tenth.

Schach [20]

Answer: 131.1287 square mm (approx)

Step-by-step explanation:

The area of a triangle,

$A=\frac{1}{2} \times s_1\times s_2\times sin \theta$

Where $s_1$ and $s_2$ are adjacent sides and $\theta$ is the include angle of these sides,

Here PR and QR are adjacent sides and ∠R is the included angle of these sides,

Thus, we can write,

$s_1 = PR= 23\text{ mm}$ , $s_2=QR=39\text{ mm}$ and $\theta = 163^{\circ}$ ,

Thus, the area of the triangle PQR,

$A=\frac{1}{2} \times 23\times 39\times sin163^{\circ}$

$A=\frac{262.257419136}{2} = 131.128709568\approx 131.1287\text{ square mm}$

5 0

3 years ago

A 60-yard long drawbridge has one end at ground level. The other end is initially at an incline of 5°. During one stage of the d

AVprozaik [17]

The drawbridge will be at an incline of 10.56 degrees. That is 5.56 degrees more than it's initial incline of 5 degrees.

If you draw a picture of this situation, you will see that you can write a sine ratio with the 11 and 60 (opposite over hypotenuse)

sin(x) = 11/60 Solve the equation for x

x = 10.56

7 0

3 years ago