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3 years ago

Gina has 25 more barrettes than Holly.

Mathematics

2 answers:

larisa [96]3 years ago

8 0

Answer:

Its 26!

Step-by-step explanation:

cestrela7 [59]3 years ago

7 0

The answer is D. 22 because 47 - 25= 22

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Which of the following shows 2 + (x + 3y) rewritten using the Associative Property of Addition?

Eduardwww [97]

The answer is B. (2 + x) + 3y.

The associative change happened when they just moved the parenthesis.

Hope this helps!

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3 years ago

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mr Goodwill [35]

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Step-by-step explanation:

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3 years ago

What is the quotient (2x^2 + 12x +18 ) divided by (x +3)

katen-ka-za [31]

2(x^2+6x+9) = 2(x+3)^2

So after divided by (x+3), answer will be:

2(x+3)

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3 years ago

There are 60 inches in 5feet. How many inches are in 8 feet ?

Murljashka [212]

Answer:

Step-by-step explanation:

Feet have 12 inches each to find the amount in x amount of inches let this equation help you

X=amount of inches

y= amount of feet

X= 12·y

in other word multiply 12 by the amount of feet to get your answer

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3 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

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

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

5 0

3 years ago