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dmitriy555 [2]
3 years ago
6

Evan and Carter are making chocolate milk which mixes milk with chocolate syrup. Evan's recipe calls for 5 cups of milk for ever

y 4 teaspoons of chocolate syrup. Carter's recipe needs 3 cups of milk for every 2 teaspoons of chocolate syrup. Whose recipe has the stronger tasting chocolate milk?
Mathematics
1 answer:
Alex73 [517]3 years ago
3 0

Answer:

Evan's recipe

Step-by-step explanation:

Given that:

Evan's recipe :

5 cups of milk for every 4 teaspoons of chocolate syrup

Carter's recipe:

3 cups of milk for every 2 teaspoons of chocolate syrup.

Recipe with the stronger tasting chocolate milk:

Take the ratio of cups of milk to teaspoons of chocolate syrup:

Evan's :

5 / 4 = 1.25

Carter's :

3 /2 = 1.5

Evan's recipe has the stronger tasting chocolate milk recipe :

5 milk cup = 4 chocolate spoons

Number of chocolate milks for 3 milk cups = x in Evan's recipe

5 = 4

3 = x

5x = 12

x = 12/5

x = 2.4 teaspoons

This is 2.4 spoons compared Carter's recipe which uses 2 spoons for every 3 milk cups

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Answer:

1. 7

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Step-by-step explanation:

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Answer:

1) n = 84

2) n = 78

3) n = 24

4) n = 16

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Step-by-step explanation:

To solve each proportion, we apply cross multiplication.

Question 1:

\frac{5}{12} = \frac{35}{n}

5n = 35*12

Simplifying both sides by 5

n = 7*12 = 84

Question 2:

\frac{n}{52} = \frac{180}{120}

120n = 52*180

Simplifying both sides by 20

6n = 52*9

Simplifying by 3

2n = 52*3

Simplifying by 2

n = 26*3 = 78

Question 3:

\frac{18}{n} = \frac{21}{28}

21n = 18*28

Simplifying both sides by 7

3n = 18*4

Simplifying both sides by 3

n = 6*4 = 24

Question 4:

\frac{n}{4} = \frac{24}{6}

\frac{n}{4} = 4

n = 16

Question 5:

\frac{10}{16} = \frac{n}{56}

16n = 56*10

Simplifying by 2, both sides

8n = 56*5

Simplifying by 8, both sides

n = 7*5 = 35

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3 years ago
Write an equation for an ellipse centered at the origin, which has foci at (0,\pm\sqrt{63})(0,± 63 ​ )left parenthesis, 0, comma
steposvetlana [31]

Answer:

\frac{x^{2} }{4312 } + \frac{y^{2} }{8281 }

Step-by-step explanation:

Since the foci are at(0,±c) = (0,±63) and vertices (0,±a) = (0,±91), the major axis is the y- axis. So, we have the equation in the form (with center at the origin) \frac{x^{2} }{b^{2} } + \frac{y^{2} }{a^{2} }.

We find the co-vertices b from b = ±√(a² - c²) where a = 91 and c = 63

b = ±√(a² - c²)

= ±√(91² - 63²)

= ±√(8281 - 3969)

= ±√4312

= ±14√22

So the equation is

\frac{x^{2} }{(14\sqrt{22}) ^{2} } + \frac{y^{2} }{91^{2} } = \frac{x^{2} }{4312 } + \frac{y^{2} }{8281 }

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Answer:

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Step-by-step explanation:

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