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Alona [7]
2 years ago
15

What is 2x+3=3x-6 I need help I have 2 hours

Mathematics
2 answers:
beks73 [17]2 years ago
6 0

Answer:

9

Step-by-step explanation:

2x+3=3x-6

6+3 = 3x - 2x

x = 9

I hope im right!!

daser333 [38]2 years ago
4 0
Answer: 9
3+6
2x+9=3x
3x-2x
9=1x
9/1= 9
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In a circle, what does the ratio A/r^2 represent?
Pepsi [2]
The ratio A/r² would represent \pi which is the circumference of a circle divided by its diameter.

You can see this because the area of a circle is \pir² and if you divided that by the radius squared, you are given pi which is the circumference of a circle divided by its diameter.
5 0
2 years ago
Evaluate 0.3y + y/z when y = 10 and z = 5
maxonik [38]

Answer:

I think the answer is 5

Step-by-step explanation:

3+2=5

4 0
3 years ago
(4x-4)+(3x-2)+(2x+6)=180
Ahat [919]

Answer:

x=20

Step-by-step explanation:

Let's solve your equation step-by-step.

4x−4+3x−2+2x+6=180

Step 1: Simplify both sides of the equation.

4x−4+3x−2+2x+6=180

4x+−4+3x+−2+2x+6=180

(4x+3x+2x)+(−4+−2+6)=180(Combine Like Terms)

9x=180

9x=180

Step 2: Divide both sides by 9.

9x

9

=

180

9

x=20

Answer:

x=20

6 0
2 years ago
Read 2 more answers
Find two positive numbers such that the sum of the first and twice the second is 88
Juli2301 [7.4K]

the 2 numbers are 8 and 40

40*2 = 80 +8 = 88

7 0
3 years ago
Whats the roots of the following three problems:
klemol [59]

QUESTION 1

The given function is

f(x)=\frac{(x-3)(x-1)}{(x-1)(x+2)}

We simplify to get;

f(x)=\frac{x-3}{x+2}

This function is equal to zero when x-3=0

\Rightarrow x=3

QUESTION 2

The given function is

f(x)=\frac{5x^2-10x+5}{2x^2-5x+3}

f(x)=\frac{5(x^2-2x+1)}{2x^2-5x+3}

We factor to get;

f(x)=\frac{5(x-1)^2}{(2x-3)(x-1)}

f(x)=\frac{5(x-1)}{(2x-3)}

This function equals zero when

5(x-1)=0

x=1

QUESTION 3

The given function is

f(x)=\frac{x^3-8}{x^2-6x+8}

We factor to get,

f(x)=\frac{(x-2)(x^2+2x+4)}{(x-2)(x+4)}

f(x)=\frac{x^2+2x+4}{x+4}

The function equals zero when x^2+2x+4=0

D=b^2-4ac

D=2^2-4(1)(4)

D=-12

Hence the equation has no real roots.

The complex roots are

x=-\sqrt{3}i-1 or x=-1+\sqrt{3}i

4 0
3 years ago
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