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anastassius [24]
3 years ago
9

a recipe calls for 6 cups of brown sugar for every 2 cups of white sugar. How many cups of brown sugar is required for every cup

of white sugar ​
Mathematics
1 answer:
sammy [17]3 years ago
3 0

Answer:

You need 3 cups of brown sugar for every 1 cup of white sugar.

Step-by-step explanation:

To find the smallest amount possible, divide both numbers by the smaller number. So, 6 (the amount of brown sugar) Divided by 2 equals 3, and 2 (the amount of white sugar) Divided by 2 equals 1.

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olasank [31]

Answer:

14

Step-by-step explanation:

Even though the triangle is "upside down"

a = (1/2)bh

b is QB = 4

h is QA = 7

a = (1/2) * 4 * 7

a = 14

8 0
3 years ago
Find the equation of the
kolbaska11 [484]

Answer:

y=-1/3 x +2/3

Step-by-step explanation:

all steps are in picture

4 0
3 years ago
Which represents where f(x) = g(x)?
Lyrx [107]

The <em><u>correct answer</u></em> is:

f(2) = g(2) and f(0) = g(0)

Explanation:

If f(x) = g(x), this means the values of the functions are the same when using the same numbers.  That means if we use 2 for x, then f(2) would need to be equal to g(2); if we use 0 for x, then f(0) would need to be equal to g(0).  This is the case in this option, so this option is correct.

3 0
3 years ago
Read 2 more answers
Y=-x^2+2x+10<br> y=x+2<br><br> Substitution <br> Please show your work<br> Need ASAP
erik [133]

Answer:

The solutions of the system of equations are the points

(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})  

(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})  

Step-by-step explanation:

we have

y=-x^{2} +2x+10 ----> equation A

y=x+2 ----> equation B

Solve the system by substitution

substitute equation B in equation A

x+2=-x^{2} +2x+10

solve for x

-x^{2} +2x+10-x-2=0

-x^{2} +x+8=0

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}

in this problem we have

-x^{2} +x+8=0

so

a=-1\\b=1\\c=8

substitute in the formula

x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}

x=\frac{-1\pm\sqrt{33}} {-2}

x=\frac{-1+\sqrt{33}} {-2}  -----> x=\frac{1-\sqrt{33}} {2}  

x=\frac{-1-\sqrt{33}} {-2}  -----> x=\frac{1+\sqrt{33}} {2}  

<em>Find the values of y</em>

For x=\frac{1-\sqrt{33}} {2}  

y=x+2

y=\frac{1-\sqrt{33}} {2}+2  ---->y=\frac{5-\sqrt{33}} {2}  

For x=\frac{1+\sqrt{33}} {2}  

y=x+2

y=\frac{1+\sqrt{33}} {2}+2  ---->y=\frac{5+\sqrt{33}} {2}  

therefore

The solutions of the system of equations are the points

(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})  

(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})  

5 0
3 years ago
What is 2/9n-5=2+5/9n<br> Btw its solving equations with variables on both sides
vova2212 [387]

Answer:

n=-0.04761

Step-by-step explanation:

given 2\div 9n-5=2+5\div 9n

2\div 9n-5\div 9n=2+5

\frac{2-5}{9n}=7

-3=7\times 9n

63n=-3

n=\frac{-3}{63}=-0.04761

n=-0.04761 answer

4 0
3 years ago
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