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3 years ago

Can someone please help me ASAP :)

Mathematics

1 answer:

jeyben [28]3 years ago

4 0

Did this a couple years ago, im kinda lost xD . sorry

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If f(x) = 3x – 12, what is f(2)?

fredd [130]

It would be B. -6.

f(x) = 3x-12
f(2) = 3(2)-12
f(2) = 6-12
f(2) = -6

5 0

3 years ago

Read 2 more answers

I need help on this question please help me

Anastasy [175]

Answer:

Step-by-step explanation:

6*[4-(42-63)]=

6*[4-(-21)]=

6*[4+21]=

6*25=150.

6*[4-9÷3]=

6*[4-27]=

6*(-23)=-138.

6*[4-3]=

6*1=6.

6 0

2 years ago

98 point reward if you help

Dennis_Churaev [7]

Answer:

the right answer is -3

Step-by-step explanation:

When we analyze the sequence it is observed that all numbers are multiplied by the value of -3.

using the formula

q = $\frac{t2}{t1}= \frac{6}{-2}= -3$

4 0

3 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

What is the volume to number 5?

almond37 [142]

It is 4200 cm hope i helped

8 0

3 years ago