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3 years ago

HELP ME PLEASEE ASAP!!!! PLEASEE

Mathematics

1 answer:

aniked [119]3 years ago

8 0

Answer:

in my view B is a answer

Step-by-step explanation:

cross section area=l×b

=5×5

=25 inch square. .

volume=A×h

=25×3

=75 inch cube..

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Can someone please help me

AVprozaik [17]

Answer:

Lines RQ and SP are perpendicular to SR

Step-by-step explanation:

SR are parallel to PQ so that means that RQ and SP are perpendicular to SR

6 0

3 years ago

Find the volume of the sphere.

nataly862011 [7]

Answer:

Volume of a sphere = 4πr³/3

r=10

V = 4πx10³/3

=4000π/3

V= 1333.33π (in terms of pi)

V= 4188.79(Nearest Hundredth).

7 0

2 years ago

Read 2 more answers

Eric, Cam, and Brooke are playing mini golf on a computer. The game stores objects in terms of x xx- and y yy-coordinates. The g

svetlana [45]

Answer:

Its Eric 9.8m

Step-by-step explanation:

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3 years ago

This table shows the scores that Jack had on his last five science tests.

arlik [135]

Answer:

the answer is B the mean can't be greater than the median

8 0

2 years ago

9. A circle has an arc of length 56pi that is intercepted by a central angle of 120 degrees. What is the radius of the circle?

SSSSS [86.1K]

Answer:

Part 4) $r=84\ units$

Part 9) $sin(\theta)=-\frac{\sqrt{5}}{3}$

Part 10) $sin(\theta)=-\frac{9\sqrt{202}}{202}$

Step-by-step explanation:

Part 4) A circle has an arc of length 56pi that is intercepted by a central angle of 120 degrees. What is the radius of the circle?

we know that

The circumference of a circle subtends a central angle of 360 degrees

The circumference is equal to

$C=2\pi r$

using proportion

$\frac{2\pi r}{360^o}=\frac{56\pi}{120^o}$

simplify

$\frac{r}{180^o}=\frac{56}{120^o}$

solve for r

$r=\frac{56}{120^o}(180^o)$

$r=84\ units$

Part 9) Given cos(∅)=-2/3 and ∅ lies in Quadrant III. Find the exact value of sin(∅) in simplified form

Remember the trigonometric identity

$cos^2(\theta)+sin^2(\theta)=1$

we have

$cos(\theta)=-\frac{2}{3}$

substitute the given value

$(-\frac{2}{3})^2+sin^2(\theta)=1$

$\frac{4}{9}+sin^2(\theta)=1$

$sin^2(\theta)=1-\frac{4}{9}$

$sin^2(\theta)=\frac{5}{9}$

square root both sides

$sin(\theta)=\pm\frac{\sqrt{5}}{3}$

we know that

If ∅ lies in Quadrant III

then

The value of sin(∅) is negative

$sin(\theta)=-\frac{\sqrt{5}}{3}$

Part 10) The terminal side of ∅ passes through the point (11,-9). What is the exact value of sin(∅) in simplified form?

see the attached figure to better understand the problem

In the right triangle ABC of the figure

$sin(\theta)=\frac{BC}{AC}$

Find the length side AC applying the Pythagorean Theorem

$AC^2=AB^2+BC^2$

substitute the given values

$AC^2=11^2+9^2$

$AC^2=202$

$AC=\sqrt{202}\ units$

$sin(\theta)=\frac{9}{\sqrt{202}}$

simplify

$sin(\theta)=\frac{9\sqrt{202}}{202}$

Remember that

The point (11,-9) lies in Quadrant IV

then

The value of sin(∅) is negative

therefore

$sin(\theta)=-\frac{9\sqrt{202}}{202}$

5 0

3 years ago