What is the density of a material, if 132 grams of the material occupies 3 cubic centimeters?

1 answer:

5 0

You need to divide here, so 132/3 = 44, so that means it's D, 44 g/cm3. Hope this helped! Brainliest is always appreciated.

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Need help please somebodt

Nostrana [21]

A≈21.46
First you find the area of the square

A=a^2 which is
=10^2
A=100

Then you find the area of the circle
A=πr^2
= π(5)^2
A≈78.54

Now you subtract the 78.54 from 100 to get 21.46

5 0

3 years ago

A right triangle is graphed on a coordinate plane. Find the length of the hypotenuse. Round your answer to the nearest tenth.

Hoochie [10]

Answer:

7.6

Step-by-step explanation:

From the graph, we know the lengths of two legs.

a = 3

b = 7

Knowing this, we can use the Pythagorean Theorem to solve for c, the hypotenuse. (remember, a^2 + b^2 = c^2)

c = √a2 + √b2

= √(7)2 + √(3)2

= √49 + √9

= √58

= 7.6

8 0

3 years ago

Use the given information to find the exact value of the trigonometric function

eimsori [14]

$\begin{gathered} \csc \theta=-\frac{6}{5} \\ \tan \theta>0 \\ \cos \frac{\theta}{2}=\text{?} \end{gathered}$

Half Angle Formula

$\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}$ $\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}$ $\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}$ $\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\ \\ \end{gathered}$

Answer:

$\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}$

Checking:

$\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{ (3rd quadrant)} \end{gathered}$

Also,