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dangina [55]
2 years ago
8

How many solutions does the equation 4x + 8 = −2 − 2 + 5x have?

Mathematics
2 answers:
Annette [7]2 years ago
7 0

Answer:

It has one solution

Step-by-step explanation:

4x + 8 = −2 − 2 + 5x

collect like terms

if + 5x goes to the left sides it become - 5x and if + 8 goes to the right sides it becomes - 8

4x - 5x  = -2-2-8

add the values on both sides

-x = - 12

divide both sides by -1

x = 12

It has one solution. The only solution is if x = 12.

Anna71 [15]2 years ago
3 0
<span>4x + 8 = −2 − 2 + 5x
5x - 4x = 8 + 4
x = 12

There's only one solution, x = 12</span>
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6. 50,000 7. 900 8. 70,000,000 9. 30,000,000

Step-by-step explanation:

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3 0
3 years ago
A circle has a diameter of 12 units, and its center lies on the x-axis. What could be the equation of the circle?
ycow [4]

Answer:

x^{2} +y^{2} =36

Step-by-step explanation:

A circle centered at (0,0) with radius r is x^{2} +y^{2} =r^{2}

Since your circle has diameter of 12, then its radius is 6.  Then r^{2} =36

So a possible answer is:  x^{2} +y^{2} =36

If you want to move the location of the center, but keep it on the x-axis, then add or subtract a number to the x, such as this:

(x+4)^{2} +y^{2} =36

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8 0
3 years ago
Find a polynomial f(x) of degree 5 that has the following zeros.
lara [203]

Answer:

Step-by-step explanation:

F(x) = (x - 2)(x - 9)(x + 4)²(x + 5)

4 0
2 years ago
Only t circled ones somebody plz help I am gonna get a 0 plz help I will give you brainleist
Vikki [24]

Step-by-step explanation:

1) A number <em>m</em> is at least four

<em>m ≥ 4  ['at least' = ≥ symbol]</em>

<em />

2) 4x  ≤ 20x; x=2

4(2) ≤ 20(2)

8 ≤ 40

[Im not sure if this is a solution or and inequality :/ ]

3)  x < 5

--> On the line, draw the number 0, and the number five. 'x < 5' also means 'x is less that five.' So, draw a dot over the line before 5, and make an arrow going to the left of the page,

<-----------------------------------o   5

<---|---|---|---|---|---|---|---|---|---|---|---|---|---|--->

                          0

Hopefully this helps! :)

4 0
3 years ago
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