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mario62 [17]
3 years ago
9

I NEED HELP ON THIS CONSTRUCTED RESPONSE! :(

Mathematics
1 answer:
Simora [160]3 years ago
7 0

Answer:

<em>x = -10</em>

Step-by-step explanation:

(9/2)(8 - x) + 36 = 102 - (5/2)(3x + 24)

Multiply both sides by 2.

9(8 - x) + 2 * 36 = 2 * 102 - 5(3x + 24)

Distribute on both sides.

72 - 9x + 72 = 204 - 15x - 120

Combine like terms on each side.

144 - 9x = 84 - 15x

Subtract 144 from both sides. Add 15x to both sides.

6x = -60

Divide both sides by 6.

x = -10

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Please help! 15 points! <br><br> Btw, you cannot answer in the file.....that's just a pic.....
olchik [2.2K]
You can find the value of x using the trigonometric function of cosine.

cos x = adj/hyp
cos 49 = x/55
55(cos 49) = x 
36.083 = x 

Hope I am right :)
8 0
3 years ago
Solve the equation-2/3x = -12 show work plz
Lorico [155]

Answer:

<h2>x = 1/18</h2>

Step-by-step explanation:

-2/3x = -12

-2 = 3x × -12

-2 = -36x

2 = 36x

2/36 = x

1/18 = x

3 0
2 years ago
Read 2 more answers
on our way to visit her parents Jennifer Drive 265 miles in 5 hours what is the average rate of speed in miles per hou
Setler79 [48]
The average speed would be 53 miles per hour, hope this helped, if you need me to explain then here is the way, so all you do is divide 265 by 5 and what I did to help make it faster was split 265 into 250 and 15 so 250 divided by 5 is 50 and 15 divided by 5 is 3 so 50 + 3 = 53!
6 0
3 years ago
Evaluate [(40 + 5) − 32] ÷ 9 ⋅ 2.
CaHeK987 [17]
(45 - 32) / 9 * 2
13 / 18

the. answer is 13 / 18
3 0
3 years ago
Read 2 more answers
1) On a standardized aptitude test, scores are normally distributed with a mean of 100 and a standard deviation of 10. Find the
Musya8 [376]

Answer:

A) 34.13%

B)  15.87%

C) 95.44%

D) 97.72%

E) 49.87%

F) 0.13%

Step-by-step explanation:

To find the percent of scores that are between 90 and 100, we need to standardize 90 and 100 using the following equation:

z=\frac{x-m}{s}

Where m is the mean and s is the standard deviation. Then, 90 and 100 are equal to:

z=\frac{90-100}{10}=-1\\ z=\frac{100-100}{10}=0

So, the percent of scores that are between 90 and 100 can be calculated using the normal standard table as:

P( 90 < x < 100) = P(-1 < z < 0) = P(z < 0) - P(z < -1)

                                                =  0.5 - 0.1587 = 0.3413

It means that the PERCENT of scores that are between 90 and 100 is 34.13%

At the same way, we can calculated the percentages of B, C, D, E and F as:

B) Over 110

P( x > 110 ) = P( z>\frac{110-100}{10})=P(z>1) = 0.1587

C) Between 80 and 120

P( 80

D) less than 80

P( x < 80 ) = P( z

E) Between 70 and 100

P( 70

F) More than 130

P( x > 130 ) = P( z>\frac{130-100}{10})=P(z>3) = 0.0013

8 0
3 years ago
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