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Dmitrij [34]
3 years ago
5

What is the solution for this eqaution:-5x – 25 = 5x + 25​

Mathematics
1 answer:
Kisachek [45]3 years ago
3 0

Answer:

x = -5

Step-by-step explanation:

Start by adding the 5x to each side of the equation to the left to get...

-25 = 5x + 25 + 5x

Then you would add 25 to each side of the equation to get...

0 = 10x + 25 + 25

Adding on, you would simplify this step to get...

0 = 10x +50

You would now have to make the x variable on one side alone so you would subtract 50 from both sides of the equation again to get...

-50 = 10x

Finally you would divide both sides of the equation by 10 to get x alone...

-50/10 = 10x/10

And simplify that to get...

x = -5

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I don’t really get this question so if anyone knows it please help
Mars2501 [29]

Answer:

116.82\:\text{square inches}

Step-by-step explanation:

The window consists of a large square and 4 smaller triangles on the outside. The side length of this square is marked as 10, and therefore the area of this square is 10^2=100\:\mathrm{in^2}.

The area of a triangle is given by \frac{1}{2}\cdot b\cdot h. Since the problem states that the overlapping squares are congruent, both legs of the triangle will be 2.9 and intersect at a 90^{\circ} angle.

Thus, the sum of all four triangles is \frac{1}{2}\cdot 2.9\cdot 2.9\cdot 4=16.82\:\mathrm{in^2}.

Therefore, the area of the window is:

100+16.82=\boxed{116.82\:\mathrm{in^2}}

3 0
3 years ago
Which of the following correctly uses absolute value to show the distance between –80 and 15?
scoray [572]
It should be |-80+15|=65 units because absolute value makes it positive.
4 0
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Suppose that in a large company, 30% of the computers are infected by adware. if 6 computers are randomly selected and tested, w
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It's either 1.8 or 0.9
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3 years ago
Complete the two-column proof.
babymother [125]
<span>The equation you start with is the given equation.
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3 0
3 years ago
A training field is formed by joining a rectangle and two semicircles, as shown below. The rectangle is 96m long and 74m wide.
Oduvanchick [21]

Answer:

11402.66 m^{2}

Step-by-step explanation:

The width of rectangle is the diameter of the semi-circle part

Area of one semicircle is given by \frac {0.5\pi d^{2}}{4}

Total area of semi circle will be 2\times\frac {0.5\pi d^{2}}{4}

Substituting 74 m for d and \pi as 3.14 we obtain

Total area semi-circle=2\times\frac {0.5*3.14\times 74^{2}}{4}=4298.66 m^{2}

Area of rectangle is given by the product of length and width

Rectangular area=96 m*74 m=7104 m^{2}

Total area of rectangular and semi-circles will be

4298.66 m^{2}+7104 m^{2}=11402.66 m^{2}

Therefore, area of training field is 11402.66 m^{2}

6 0
3 years ago
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