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yan [13]
3 years ago
14

Luca had a large container of yogurt with 1 and StartFraction 5 Over 8 EndFraction pounds of yogurt left in it. If a serving of

yogurt is StartFraction 3 Over 8 EndFraction of a pound, to the nearest whole serving, how many servings of yogurt are in the container?
Mathematics
1 answer:
Nataliya [291]3 years ago
3 0

Answer:There are 4 serving in the container.

Step-by-step explanation:

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A pizza company makes pizza in three different sizes: small, medium, large. There are four possible toppings: pepperoni, sausage
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total with one topping = 12

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3 years ago
When converting measurements in the metric system, you can move the decimal point to the left or to the right. Why? Select all t
Minchanka [31]

Answer:

A,C, D

Step-by-step explanation:

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3 years ago
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Describe a real world situation that can be modeled by 5t
jeka57 [31]

I used to shovel snow for $5 an hour.

If ' t ' was the number of hours I spent shoveling somebody's driveway,
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2 years ago
Due tomorrow Please help!!
anastassius [24]

Answer:

A

\frac{1}{8} = \frac{b}{4^{2} }

b = \frac{1}{8} × 4^{2}

b = \frac{4^{2} }{8}

then b = \frac{c^{2} }{a} that is ORANGE

B

-2 - v = -7

v = -7 +2

then v = k + m that is BROWN

C

\frac{14}{q} = \frac{-7}{-4}

q = \frac{14 * -4 }{-7}

then q = \frac{ps}{r} that is YELLOW

D

4m + 2(n) = 5 (n)

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4m = 3n

m = \frac{3n}{4}

then m =  \frac{3n}{4} that is RED

E

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\frac{x}{-5} = -8 + 6

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7 0
3 years ago
The corners of a meadow are shown on a coordinate grid. Ethan wants to fence the meadow. What length of fencing is required?
Nuetrik [128]

Answer:

34.6 units

Step-by-step explanation:

The lenght of fencing required is the total distance between point A to B, B to C, C to D, and D to A. That is the distance between all 4 corners of the meadow.

The coordinates of the corners of the meadow is shown on a coordinate plane in the attachment. (See attachment below).

Let's use the distance formula to calculate the distance between the 4 corners of the meadow using their coordinates as follows:

Distance between point A(-6, 2) and point B(2, 6):

AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

A(-6, 2)) = (x_1, y_1)

B(2, 6) = (x_2, y_2)

AB = \sqrt{(2 - (-6))^2 + (6 - 2)^2}

AB = \sqrt{(8)^2 + (4)^2}

AB = \sqrt{64 + 16} = \sqrt{80}

AB = 8.9 (nearest tenth)

Distance between B(2, 6) and C(7, 1):

BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

B(2, 6) = (x_1, y_1)

C(7, 1) = (x_2, y_2)

BC = \sqrt{(7 - 2)^2 + (1 - 6)^2}

BC = \sqrt{(5)^2 + (-5)^2}

BC = \sqrt{25 + 25} = \sqrt{50}

BC = 7.1 (nearest tenth)

Distance between C(7, 1) and D(3, -5):

CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

C(7, 1) = (x_1, y_1)

D(3, -5) = (x_2, y_2)

CD = \sqrt{(3 - 7)^2 + (-5 - 1)^2}

CD = \sqrt{(-4)^2 + (-6)^2}

CD = \sqrt{16 + 36} = \sqrt{52}

CD = 7.2 (nearest tenth)

Distance between D(3, -5) and A(-6, 2):

DA = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

D(3, -5) = (x_1, y_1)

A(-6, 2) = (x_2, y_2)

DA = \sqrt{(-6 - 3)^2 + (2 - (-5))^2}

DA = \sqrt{(-9)^2 + (7)^2}

DA = \sqrt{81 + 49} = \sqrt{130}

DA = 11.4 (nearest tenth)

Length of fencing required = 8.9 + 7.1 + 7.2 + 11.4 = 34.6 units

8 0
3 years ago
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