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Galina-37 [17]
3 years ago
12

Can someone help me on this?

Mathematics
2 answers:
Mamont248 [21]3 years ago
5 0
A = 89.2


Hope it’s right sorry if it isn’t.
RoseWind [281]3 years ago
4 0

\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h=height\\ a,b=\stackrel{parallel~sides}{bases}\\[-0.5em] \hrulefill\\ h=8\\ a=15.7\\ b=6.6 \end{cases}\implies A=\cfrac{8(15.7+6.6)}{2}\implies A=89.2~in^2

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Suppose a charity received a donation of $22.2 million. If this represents 34% of the charity’s donated funds, what is the total
Novosadov [1.4K]

Answer:

ok so lets make this into a algrebra equation

22.2(mil)=0.34x

divide by 0.34

65294117.6471=x

x=65294118

Hope This Helps!!!

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2 years ago
Which ratio correctly compares 12 yd to 20 ft, when the lengths are written using the same units?
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The believe that the best answer among the choices provided by the question is the second choice "5 : 3".Hope my answer would be a great help for you.    If you have more questions feel free to ask here at Brainly.
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3 years ago
Alexa is using 6 pounds of apples to make pies.
bixtya [17]

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96

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3 years ago
Find two positive real numbers whose product is a maximum. the sum is 80
Eduardwww [97]
Let's say the numbers are "a" and "b"

thus \bf y=ab\qquad 
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4 0
3 years ago
Read 2 more answers
Towns P,Q,R and S are shown. Q is 35 km due East of P S is 15km due West of P R is 15km due South of P Work out the bearing of R
Anni [7]

Answer:

Part A

The bearing of the point 'R' from 'S' is 225°

Part B

The bearing from R to Q is approximately 293.2°

Step-by-step explanation:

The location of the point 'Q' = 35 km due East of P

The location of the point 'S' = 15 km due West of P

The location of the 'R' = 15 km due south of 'P'

Part A

To work out the distance from 'R' to 'S', we note that the points 'R', 'S', and 'P' form a right triangle, therefore, given that the legs RP and SP are at right angles (point 'S' is due west and point 'R' is due south), we have that the side RS is the hypotenuse side and ∠RPS = 90° and given that \overline{RP} = \overline{SP}, the right triangle ΔRPS is an isosceles right triangle

∴ ∠PRS = ∠PSR = 45°

The bearing of the point 'R' from 'S' measured from the north of 'R' = 180° + 45° = 225°

Part B

∠PRQ = arctan(35/15) ≈ 66.8°

Therefore the bearing  from R to Q = 270 + 90 - 66.8 ≈ 293.2°

6 0
3 years ago
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