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ivann1987 [24]
2 years ago
5

Abygail is building a dollhouse. She has boards that are two different lengths. One long board is 10 inches longer than the tota

l length of three of the short boards.
If one of the long boards is 37 inches long, how long is a short board?
Mathematics
1 answer:
nikitadnepr [17]2 years ago
7 0

Answer:

9 in

Step-by-step explanation:

a short bored is 9 in

37-10=27

27÷3=9

So that means a short bored is 9 in

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Put these fractions in order of size, smallest to largest.<br> 7/10<br> 2/3<br> 4/5<br> 11/15
Alla [95]

Answer:

11/15 7/10 4/5 2/3

Step-by-step explanation:

the bigger the fraction the smaller the number

4 0
2 years ago
Find the area of the circle. Diameter = 12 mi<br><br> answer choice:<br> 36π mi<br> 144π mi
elena-s [515]

Answer:

36π mi

Step-by-step explanation:

Area of a circle = πr²

Diameter = 12 mi. Therefore, radius=12/2 = 6 mi

πr²

π x (6)²

π x 36

Hence, the answer is 36π mi

3 0
3 years ago
Read 2 more answers
What is 2 7/4 in simplest form
Leona [35]

2 7/4=3.75

THE ANSWER IS 3.75

4 0
3 years ago
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Hello, could anyone help me on part A and part B
solniwko [45]

Answer:

9 answer ok like floow on

7 0
3 years ago
What is the sum of 20x^2-10x-30
ivann1987 [24]

Answer:

The sum of the roots is 0.5

Step-by-step explanation:

<u><em>The correct question is</em></u>

What is the sum of the roots of 20x^2-10x-30

we know that

In a quadratic equation of the form

ax^{2} +bx+c=0

The sum of the roots is equal to

-\frac{b} {a}

in this problem we have

20x^{2} -10x-30=0  

so

a=20\\b=-10\\c=-30

substitute

-\frac{(-10)} {20}=0.5

<u><em>Verify</em></u>

Find the roots of the quadratic equation

The formula to solve a quadratic equation is equal to

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

a=20\\b=-10\\c=-30

substitute

x=\frac{-(-10)\pm\sqrt{-10^{2}-4(20)(-30)}} {2(20)}

x=\frac{10\pm\sqrt{2,500}} {40}

x=\frac{10\pm50} {40}

x=\frac{10+50} {40}=1.5

x=\frac{10-50} {40}=-1

The roots are x=-1 and x=1.5

The sum of the roots are

-1+1.5=0.5 ----> is ok

5 0
3 years ago
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