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brilliants [131]
3 years ago
5

What is the density of the brown sugar if an 8cm3 block of brown sugar is 12.9 g

Mathematics
1 answer:
julia-pushkina [17]3 years ago
4 0
The answer is 1.6125 g/cm3
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Madelyn's car used 6 gallons of gas to drive 126 miles. At what rate does her car use
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Answer:

Step-by-step explanation:

You are given    GALLONS     and   MILES

   you want to find MILES / GALLON

     so just put those numebrs in

         MILES / GALLON =  126 miles / 6 gallons = 21 m/g  (or 21 MPG)

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the item would cost $9000

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Which of the following is less than 0.625?<br>A. 0.623<br>B. 0.651<br>C. 0.627<br>D. 1.624<br><br>​
galina1969 [7]

Answer:

A. 0.623

Step-by-step explanation:

Hope it helps you in your learning process.

3 0
3 years ago
Read 2 more answers
What is the answer for this
e-lub [12.9K]

Answer:

A

Step-by-step explanation:

The green is the starting place, when you move from a point on the green triangle to the blue one, you go 5 to the left and 3 down.

8 0
3 years ago
Match each equation with its solution set. Tiles a2 − 9a + 14 = 0 a2 + 9a + 14 = 0 a2 + 3a − 10 = 0 a2 + 5a − 14 = 0 a2 − 5a − 1
sattari [20]
We have that

N 1)
a²<span> − 9a + 14 = 0 
</span>

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² − 9a)=-14

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² − 9a+20.25)=-14+20.25

Rewrite as perfect squares

(a-4.5)²=6.25--------> (a-4.5)=(+/-)√6.25

a1=4.5+√6.25-----> a1=7

a2=4.5-√6.25-----> a2=2

the solution problem N 1 is the pair {7, 2}


N 2) 

a²<span> + 9a + 14 = 0
</span>

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² + 9a)=-14

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² +9a+20.25)=-14+20.25

Rewrite as perfect squares

(a+4.5)²=6.25--------> (a+4.5)=(+/-)√6.25

a1=-4.5+√6.25-----> a1=-2

a2=-4.5-√6.25-----> a2=-7

the solution problem N 2 is the pair {-2,-7}

N 3) 

a² + 3a − 10 = 0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² + 3a)=10

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² + 3a+2.25)=10+2.25

Rewrite as perfect squares

(a+1.5)²=12.25------> (a+1.5)=(+/-)√12.25

a1=-1.5+√12.25-----> a1=2

a2=-1.5-√12.25-----> a2=-5

the solution problem N 3 is the pair {2, -5}


N 4)

a²<span> + 5a − 14 = 0
</span>

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² + 5a) =14

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² + 5a+6.25) =14+6.25

Rewrite as perfect squares

(a+2.5)² =20.25-------> (a+2.5)=(+/-)√20.25

a1=-2.5+√20.25-----> a1=2

a2=-2.5-√20.25-----> a2=-7

the solution problem N 4 is the pair {2, -7}


N 5) 

a² − 5a − 14 = 0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(a² − 5a)=14

Complete the square  Remember to balance the equation by adding the same constants to each side 

(a² − 5a+6.25)=14+6.25

Rewrite as perfect squares

(a-2.5)²=2025--------> (a-2.5)=(+/-)√20.25

a1=2.5+√20.25-----> a1=7

a2=2.5-√20.25-----> a2=-2

the solution problem N 5 is the pair {7, -2}

3 0
3 years ago
Read 2 more answers
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