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3 years ago

8

HELP ME

1 answer:

Sophie [7]3 years ago

5 0

Answer:

sin 42 = h/400

Step-by-step explanation:

You might be interested in

2 2/3 multiplied by 5 4/5

grin007 [14]

$2 \frac{2}{3} = \frac{8}{3} 
$
$5 \frac{4}{5} = \frac{29}{5}$
$\frac{8}{3} * \frac{29}{5} =X$
Solve for X. What do you think the answer is based on that?

8 0

3 years ago

Substitution for:<br> X-5y=10<br> 2x-10y=20

sergey [27]

Since you need an isolated variable to use the substitution method, we need to re-arrange one of the equations. This will probably be easiest to do with the first one.
Add 5y to both sides of the first equation.
x=10+5y
Now, in the second equation, put in 10+5y in any spot that has an x.
2(10+5y)-10y=20
Distribute the 2 to both numbers in the parenthesis.
20+10y-10y=20
Combine like terms.
20=20
This means that the two equations are actually the same. You can see this if you multiply the whole first equation by 2
2(x-5y=10)
2x-10y=20, which is the same as the second equation. Therefore, the two equations are actually the same one.

5 0

3 years ago

10. Which quadrilateral is not a parallelogram?

babunello [35]

Idek broooooooooo good luck tho

8 0

2 years ago

Read 2 more answers

. The table shows the rainfall (in inches) for three

swat32

Answer:14 the apple

Step-by-step explanation:

3 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

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

$AC^2=202$

$AC=\sqrt{202}\ units$

so

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

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