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iren2701 [21]
3 years ago
5

Sarah mixes 2/3 cup of low fat milk and 1/2 cup of skim milk and the mixture has 111 calories. She knows that if she mixes one c

up of low fat milk and one cup of skim milk that the mixture will have 188 calories. How many calories are in one cup of low fat milk and how many calories are in
one cup of skim milk?

A 102 calories in a cup of low fat milk and 86 calories in a cup of skim milk
B 102 calories in a cup of low fat milk and 40 calories in a cup of skim milk
C 86 calories in a cup of low fat milk and 102 calories in a cup of skim milk
D 40 calories in a cup of low fat milk and 102 calories in a cup of skim milk

Mathematics
1 answer:
pickupchik [31]3 years ago
7 0

The calories in a cup of low fat milk is 102 calories and the  calories in a cup of milk in skim 86 calories .

<h3>What are the two simultaneous equations that can be sued to represent this question?</h3>

2/3l + 1/2s = 111

4l + 3s = 666 equation 1

l + s = 188 equation 2

Where:

l = calories in the low fat milk

s = calories in the skim milk

<h3>What is the calories in the skim milk?</h3>

Multiply equation 2 by 4

4l + 4s = 752 equation 3

Subtract equation 1 from 3

s = 86

<h3>What is the calories in the low fat milk?</h3>

Substitute for s in equation 2

l + 86 = 188

l = 188 - 86

l = 102

To learn more about simultaneous equations, please check: brainly.com/question/25875552

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In ΔUVW, w = 44 cm, u = 83 cm and ∠V=141°. Find the area of ΔUVW, to the nearest square centimeter.
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Answer:

Area  \approx 1149\ cm^2

Step-by-step explanation:

<u>Given that:</u>

ΔUVW,

Side w = 44 cm, (It is the side opposite to \angle W)

Side u = 83 cm (It is the side opposite to \angle U)

and ∠V=141°

Please refer to the attached image with labeling of the triangle with the dimensions given.

Area of a triangle with two sides given and angle between the two sides can be formulated as:

A = \dfrac{1}{2}\times a\times b\times sinC

Where a and b are the two sides and

\angle C is the angle between the sides a and b

Here we have a = w = 44cm

b = u = 44cm

and ∠C= ∠V=141

Putting the values to find the area:

A = \dfrac{1}{2}\times 44\times 83\times sin141\\\Rightarrow A = \dfrac{1}{2} \times 3652 \times sin141\\\Rightarrow A =1826 \times 0.629\\\Rightarrow A  \approx 1149\ cm^2

So, the <em>area </em>of given triangle to the nearest square centimetre is:

Area  \approx 1149\ cm^2

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