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3 years ago

a recipe cause for 4 cups of flour and 1 1/4 cups of milk how much milk do you use for 2 cups of flour

Mathematics

2 answers:

jeyben [28]3 years ago

5 0

Answer:

5/8

Step-by-step explanation:

4 cups of flour=1 1/4 cups of milk

4=5/4

Dividing both sides by 2

4/2=(5/4)/2

2 cups of flour=5/8 cups of milk

nignag [31]3 years ago

3 0

Answer:

Step-by-step explanation:

Ratio of flour and milk = 4 : 1 1/4 = 4 : 5/4

4/2 : (5/4)/2 = 2 : 5/4*2 = 2: 5/8

5/8 cup of milk

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2. The rectangular floor of a garage has an area of 198 square feet. Andy knows that the floor is 7

Charra [1.4K]

Answer:

length = 18 ft

Step-by-step explanation:

Let w = width

L = w+7

Area = l*w

198 = (w+7) (w)

Distribute

198 = w^2 +7w

Subtract 198 from each side

0 = w^2 +7w -198

Factor

What 2 numbers multiply to -198 and add to 7

18* -11 = -198

18+-11 = 7

0=(w+18) (w-11)

Using the zero product property

w+18 =0 w-11=0

w= -18 w=11

We cannot have negative width so w=11

We want the length

l = w+7

l = 11+7 = 18

6 0

2 years ago

The functions f(x) and g(x) are defined below.

erica [24]

-3 is you answer :)
:^

7 0

3 years ago

A function follows the rule y = -75 - 5 x . When the function's output is 25, the equation is 25 = -75 - 5 x .

Alex787 [66]

Answer:

The function's input = x = -20

Step-by-step explanation:

Given the function

y = -75 - 5x

Given that the output = y = 25

substituting the value y=25 and solve to find the input 'x'

25 = -75 - 5x

5x = -75-25

5x = -100

Divide bothe side by 5

x = -20

Thus, the function's input = x = -20

5 0

3 years ago

Simplify this algebraic equation: 3y - 7(5x + 2y)

Taya2010 [7]

Answer: -11y - 35x

Step-by-step explanation:

This problem utilizes the Distributive Property of Multiplication. It states that: a(x + y) = ax + ay.

Thus, utilizing this property (distributing -7), change the equation into:

3y - 35x - 14y

Then, simply combine like terms to get:

-11y - 35x

Hope it helps :) and let me know if you want me to elaborate.

8 0

2 years ago

Sort the ratios listed at the right into bins so that equivalent ratios are grouped together:

cestrela7 [59]

Step $1$

Find the irreducible fraction in each ratio

case 1) $\frac{40}{64}$

Divide by $8$ boths numerator and denominator

$\frac{40}{64}=\frac{5}{8}$

case 2) $\frac{5}{75}$

Divide by $5$ boths numerator and denominator

$\frac{5}{75}=\frac{1}{15}$

case 3) $\frac{14}{35}$

Divide by $7$ boths numerator and denominator

$\frac{14}{35}=\frac{2}{5}$

case 4) $\frac{24}{64}$

Divide by $8$ boths numerator and denominator

$\frac{24}{64}=\frac{3}{8}$

case 5) $\frac{6}{15}$

Divide by $3$ boths numerator and denominator

$\frac{6}{15}=\frac{2}{5}$

case 6) $\frac{65}{104}$

Divide by $13$ boths numerator and denominator

$\frac{65}{104}=\frac{5}{8}$

case 7) $\frac{66}{176}$

Divide by $22$ boths numerator and denominator

$\frac{66}{176}=\frac{3}{8}$

case 8) $\frac{12}{30}$

Divide by $6$ boths numerator and denominator

$\frac{12}{30}=\frac{2}{5}$

case 9) $\frac{15}{225}$

Divide by $15$ boths numerator and denominator

$\frac{15}{225}=\frac{1}{15}$

case 10) $\frac{6}{16}$

Divide by $2$ boths numerator and denominator

$\frac{6}{16}=\frac{3}{8}$

case 11) $\frac{15}{24}$

Divide by $3$ boths numerator and denominator

$\frac{15}{24}=\frac{5}{8}$

case 12) $\frac{48}{128}$

Divide by $16$ boths numerator and denominator

$\frac{48}{128}=\frac{3}{8}$

Step $2$

Sort the ratios into bins

1) First Bin

$Ratio=\frac{5}{8}$

$\frac{40}{64}$

$\frac{65}{104}$

$\frac{15}{24}$

2) Second Bin

$Ratio=\frac{1}{15}$

$\frac{5}{75}$

$\frac{15}{225}$

3) Third Bin

$Ratio=\frac{2}{5}$

$\frac{14}{35}$

$\frac{6}{15}$

$\frac{12}{30}$

4) Fourth Bin

$Ratio=\frac{3}{8}$

$\frac{24}{64}$

$\frac{66}{176}$

$\frac{6}{16}$

$\frac{48}{128}$

3 0

2 years ago

Read 2 more answers