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2 years ago

I wonder who can solve this ? 20×20÷5+100-50

2 answers:

8 0

Use PEMDAS

20 * 20/5 + 100 - 50

Multiply and divide first

20 * 20 = 400

400/5 = 80

Add 100 and then subtract 50

80 + 100 = 180

180 - 50 = 130

130 is the answer.

Margaret [11]2 years ago

5 0

20 * 20/5 + 100 - 50 Multiply and divide first 20 * 20 = 400 400/5 = 80 Add 100 and then subtract 50 80 + 100 = 180 180 - 50 = 130 130 is the answer.

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Answer: 20 miles/1 gallon would be the unit rate so NO 60 miles/3 gallons is not the unit rate

Explanation:

60 / 3 = 20

I hope this helped!

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3 years ago

f. A fair coin is thrown in the air four times. If the coin lands with the head up on the first three tosses, what is the probab

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3 years ago

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Which of the following is the radical expression of 2d^7/10

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2 years ago

Y=-x^2+2x+10 y=x+2 Substitution Please show your work 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

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

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

substitute in the formula

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

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

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

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

Find the values of y

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

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

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

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2 years ago