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tester [92]
3 years ago
15

Please help me answer

Mathematics
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
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