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enot [183]
3 years ago
6

Help it was due 5 min ago :((((( (show all your work)

Mathematics
2 answers:
nevsk [136]3 years ago
7 0
X = 90
ALSO I LOVE YOUR USER AND PFP!!!!
goblinko [34]3 years ago
5 0
X is 90 because its a right angle
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he gross income of Maurice Vaughn is $785 per week. His deductions are $42.25, FICA tax; $90.33, income tax; 2% state tax; 1% ci
docker41 [41]

Answer:

A. $605.32

Step-by-step explanation:

The state tax, city tax, and retirement fund add to 6% of Vaughn's wages, or ...

... $785 · 6% = $47.10

Then after subtracting all the deductions, Vaughn's take-home pay is ...

... $785 -42.25 -90.33 -47.10 = $605.32 . . . . matches A

3 0
3 years ago
300 mi /20 sec = ? Mi /1 min
snow_lady [41]

Answer:

900

Step-by-step explanation:

To get your answer, you can just multiply 20 seconds by 3 to get 60 seconds, which is a minute.

Since whatever you do on one side you have to do to the other, multiply 300 mi by 3 as well. That would get you 900.

4 0
3 years ago
2x – 3y = 18<br> Solve for y.
Nataly [62]

Answer:

y= -6+2/3x

Step-by-step explanation:

2x-3y=18

Minus the 2x from both sides

-3y=18-2x

Then divide both sides by -1

-3y÷-1=18÷-1 -2x÷-1

Lastly divide both sides by 3

3y÷3= -18÷3+2x÷3

7 0
3 years ago
Which best describes the life cycles of moss and angiosperm?
Aleksandr-060686 [28]
It would have to be iv

4 0
3 years ago
Read 2 more answers
Find the area of the region which is inside the polar curve r=5sin(θ) but outside r=4. Round your answer to four decimal places
natka813 [3]

The area of the region which is inside the polar curve r = 5 sinθ but outside r = 4 will be 3.75 square units.

<h3>What is an area bounded by the curve?</h3>

When the two curves intersect then they bound the region is known as the area bounded by the curve.

The area of the region which is inside the polar curve r = 5 sinθ but outside r = 4 will be

Then the intersection point will be given as

\rm 5 \sin \theta  = 4\\\\\theta = 0.927 , 2.214

Then by the integration, we have

\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214}[ (5 \sin \theta)^2 - 4^2] d\theta \\\\\\\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214} [25\sin ^2 \theta - 16] d\theta \\\\\\\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214} [ \dfrac{25}{2}(1 - \cos 2\theta ) - 16] d\theta \\

\rightarrow \dfrac{1}{2} [\dfrac{25 \theta }{5} - \dfrac{25 \cos 2\theta }{2} - 16\theta]_{0.927}^{2.214} \\\\\\\rightarrow \dfrac{1}{2} [\dfrac{25(2.214 - 0.927) }{5} - \dfrac{25 (\cos 2\times 2.214 - \cos 2\times 0.927) }{2} - 16(2.214 - 0.927]\\

On solving, we have

\rightarrow \dfrac{1}{2} \times 7.499\\\\\rightarrow 3.75

Thus, the area of the region is 3.75 square units.

More about the area bounded by the curve link is given below.

brainly.com/question/24563834

#SPJ4

5 0
2 years ago
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