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1 year ago

maeve likes to make her salads in advance she made 15 salads for 3 days how many salads does she eat per day unit rate

Mathematics

1 answer:

Tju [1.3M]1 year ago

7 0

The number of salads Maeve eats per day if she made 15 salads for 3 days is 5 salads eaten per day.

Unit rate refers to the number of units of the first type of quantity corresponding to one unit of the second type of quantity.

Number of salads = 15 salads
Number of days = 3 days

Unit rate of salad eaten per day = Number of salads ÷ Number of days

= 15 ÷ 3

= 5 salads eaten per day

A salad is a food made primarily of a mixture of raw or cold ingredients, typically vegetables, usually served with a dressing such as vinegar or mayonnaise.

Therefore, the number of salads Maeve eats per day if she made 15 salads for 3 days is 5 salads eaten per day.

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A.
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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

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2 years ago

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Answer:

Step-by-step explanation:

Lines EG and CD are cut by transversal CF.

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3 years ago