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3 years ago

15

Please help me answer

2 answers:

melamori03 [73]3 years ago

4 0

Answer:

A

Step-by-step explanation:

3x=x-12

plug in 3 for x

3(3)=(3)-12

9=-9

masha68 [24]3 years ago

4 0

Answer:

Let the number be X;

From the statement, X-12=3(-X)

Collecting the like terms, X+3X =12

4X=12

Dividing both sides by 4

X=3

Step-by-step explanation:

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Help with this math question

tresset_1 [31]

The given inequality is:

$|-8x+24| \leq 16$

This inequality can be divided in two parts as:

a) $-16 \leq -8x +24$
b) $-8x + 24 \leq 16$

Solving part a:

$-16 \leq -8x+24 \\ \\ 
-40 \leq -8x \\ \\ 
5 \geq x \\ 
or \\ 
x \leq 5$

Solving part b:

$-8x+24 \leq 16 \\ \\ 
-8x \leq -8 \\ \\ 
x \geq 1$

Therefore, the solution to the given inequality is $x \leq 5$ and $x \geq 1$ . Combining both the ranges we get the solution: $1 \leq x \leq 5$ .

In interval notation, this solution can be expressed as [1,5]

3 0

3 years ago

Read 2 more answers

A and b are positive integers and a–b = 3. Evaluate the following:

yulyashka [42]

1/27

1/27

125

Step-by-step explanation:

Given that,

a - b = 3

9^(1/2b) /3^a = 3^(2/2b) /3^a

= 3^b/3^a

= 3^(b-a)

= 3^(-3)

= 27^(-1)

= 1/27

27^(1/3b) /9^(1/2a) = 3^(3/3b) /3^(2/2a)

= 3^b/3^a

= 3^(b-a)

= 3^(-3)

= 27^(-1)

= 1/27

125^(1/3a) /25^(1/2b) = 5^(3/3a) /5^(2/2b)

= 5^a/5^b

= 5^(a- b)

= 5^3

= 125

6 0

3 years ago

Find the exact area of the sector of the circle with the following radius and central angle

Lostsunrise [7]

THe area of the whole circle is pi r^2 = 6^2 * pi = 36pi

There are 360 degrees in a crcle so this sectors area i (30/360) *36pi

= 3 pi

5 0

3 years ago

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

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

so

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

substitute in the formula

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

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

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

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

<em>Find the values of y</em>

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

-29=6+5xSimplify your answer as much as possible.

Ivahew [28]

Answer:

−7

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers