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Kitty [74]
1 year ago
8

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.

<h3>Unit rate</h3>

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.

Learn more about unit rate:

brainly.com/question/4895463

#SPJ1

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goldfiish [28.3K]
A.
EXPLANATION:
The mode is the peak of the data, in this data set, 5/8
The median is the center of data, in the data set, 1/2
5/8>1/2
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6 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
Simplify 7 square root of 3 end root minus 4 square root of 6 end root plus square root of 48 end root minus square root of 54
Katen [24]
7\sqrt3-4\sqrt6+\sqrt{48}-\sqrt{54}\\\\=7\sqrt3-4\sqrt6+\sqrt{16\cdot3}-\sqrt{9\cdot6}\\\\=7\sqrt3-4\sqrt6+\sqrt{16}\cdot\sqrt3-\sqrt9\cdot\sqrt6\\\\=7\sqrt3-4\sqrt6+4\sqrt3-3\sqrt6\\\\=\boxed{11\sqrt3-7\sqrt6}
3 0
3 years ago
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elixir [45]
The answer is 6 :), I hope this helps <33!3&9/&
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3 years ago
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In the straightedge and compass construction of the parallel line below, which of the following reasons can you use to prove tha
Eva8 [605]

Answer:

D.

Step-by-step explanation:

Lines EG and CD are cut by transversal CF.

By construction, ∠FEG=∠FCD. These two angles are corresponding angles.

Since two corresponding angles are congruent, then lines EG and CD are parallel (by  converse of the corresponding angles postulate).

Converse of the Corresponding Angles Postulate: If the corresponding angles formed by two lines and a transversal are congruent, then lines are parallel.

3 0
3 years ago
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