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jarptica [38.1K]
4 years ago
12

Can you help me please?

Mathematics
1 answer:
vovangra [49]4 years ago
6 0

Answer:

D

Step-by-step explanation:

Range

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5/3 divided by (-6/7).<br>Write your answer in simplest form.<br><br>Please please Help me!! ​
lana [24]

Answer:

-1  17/18

Step-by-step explanation:

I solved this.

7 0
2 years ago
Read 2 more answers
44 88 88 66 22 99 88 77 99 66 99 55 22 66 22 99 88 77 77 99 ​(a) minequals= 22 ​(simplify your​ answer.) upper q 1 equalsq1= not
d1i1m1o1n [39]

I’m sorry. What’s the question for this problem?

5 0
3 years ago
Please find measure of angle s. PLEASE HELP FAST!!
Studentka2010 [4]

Answer:

27

Step-by-step explanation:

15x-31 = 9x+11

6x-31 = 11

6x = 42

x = 7

15x -31 + 9x + 11 + s = 180

15(7) -31 + 9(7) + 11  + s = 180

105 - 31 + 63 + 11 + s = 180

153 + s = 180

s = 27

8 0
3 years ago
The braking distance, in feet of a car a Travling at v miles per hour is given.
irakobra [83]

The braking distance is the distance the car travels before coming to a stop after the brakes are applied

a. The braking distances are as follows;

  • The braking distance at 25 mph, is approximately <u>63.7 ft.</u>
  • The braking distance at 55 mph,  is approximately <u>298.35 ft.</u>
  • The braking distance at 85 mph,  is approximately <u>708.92 ft.</u>

b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s

Reason:

The given function for the braking distance is D = 2.6 + v²/22

a. The braking distance if the car is going 25 mph is therefore;

25 mph = 36.66339 ft./s

D = 2.6 + \dfrac{36.66339^2}{22} = 63.7 \ ft.

At 25 mph, the braking distance is approximately <u>63.7 ft.</u>

At 55 mph, the braking distance is given as follows;

55 mph = 80.65945  ft.s

D = 2.6 + \dfrac{80.65945^2}{22} \approx 298.35 \ ft.

At 55 mph, the braking distance is approximately <u>298.35 ft.</u>

At 85 mph, the braking distance is given as follows;

85 mph = 124.6555 ft.s

D = 2.6 + \dfrac{124.6555^2}{22} \approx 708.92 \ ft.

At 85 mph, the braking distance is approximately <u>708.92 ft.</u>

b. The speed of the car when the braking distance is 450 feet is given as follows;

450 = 2.6 + \dfrac{v^2}{22}

v² = (450 - 2.6) × 22 = 9842.8

v = √(9842.2) ≈ 98.211 ft./s

The car was moving at v ≈ <u>98.211 ft./s</u>

Learn more here:

brainly.com/question/18591940

8 0
2 years ago
Find the interest earned.
MakcuM [25]
September
30-10=20 days
October 31
November 10
Time=20+31+10=61 days
A=42,000×(1+0.035÷365)^(61)
A=42,246.38
Interest earned
I=42,246.38−42,000
I=246.38
5 0
3 years ago
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