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3 years ago

Help me with this please!!!!

Mathematics

2 answers:

sveticcg [70]3 years ago

6 0

Answer:

option c is the correct answer

sammy [17]3 years ago

4 0

The answer is C!!! Hope this helps ❤️

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Susannah has a high-yield savings account into which she deposits $180 each month. At the beginning of the year, Susannah's acco

stiks02 [169]

A. 2860

180 x 12 (months in a year) = $2160
$2160+700= $2860

8 0

3 years ago

Read 2 more answers

Solve for x. (30/x) = 4/6

andre [41]

X=30×6/4
x=45.
hope it helps

4 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 63.7 ft.

The braking distance at 55 mph, is approximately 298.35 ft.

The braking distance at 85 mph, is approximately 708.92 ft.

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 63.7 ft.

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 298.35 ft.

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 708.92 ft.

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 ≈ 98.211 ft./s

Learn more here:

brainly.com/question/18591940

8 0

2 years ago

HELP!!!

Reptile [31]

Answer:

(a) $ 30000 + 1500 t

(b) $ 52500

Step-by-step explanation:

Initial profit = # 30,000

Profit increases every year by 5 %.

(a) Let the profit after t year is

P = $ 30,000 + 5% of 30,000 t = $ 30000 + $ 1500 t

(b) t = 15 years

P = $ 30000 + $ 1500 x 15 = $ 52500

5 0

2 years ago

The proper natation for a normal distribution with a mean of 250 and a standard deviation of 25 is?

Zolol [24]

N(250,25)
That's all it is

5 0

3 years ago