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2 years ago

BRAINLIEST AND WORTH 10 POINTS!

Mathematics

1 answer:

suter [353]2 years ago

5 0

3x = 10 because you would minus the 3 to the 13 then you would do 10 divided 3

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Need help.Pleaseeeee

Marina CMI [18]

Answer:

The first one and the last one

Step-by-step explanation:

To sub x>0 and x<0 into every functions.

Then, u will get some function that y doesn't have a negative value or positive value.

Then, these function doesn't fulfill the range.

Then, u will get the answer.

8 0

2 years ago

Two boxes of identical birthday candles are pictured: a 2-ounce box and a 6-ounce box. There are 96 candles in the large box. Ho

Lostsunrise [7]

Ratio of large box
96:6
Thus 96/6 = 16

Small box
2*16 = 32
There are 32 candles in the small box.

5 0

2 years ago

jordan has 90 photos.explain three ways he can arrange the photos in his album so he has the same number of photos on each page

erik [133]

Answer:

Divide the number of pages in the album by the 90 photos

Place one photo on a page at a time then start over when he reaches the end of the album and continue to repeat until all photos are gone

Just have one page of photos

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

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

2 years ago

What is 7/20 in decimals

FromTheMoon [43]

Answer:

0.35 :)

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers