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Licemer1 [7]
4 years ago
13

How do solve -5(-2x+7)-3=52

Mathematics
1 answer:
Viefleur [7K]4 years ago
4 0
-5(-2x+7)-3=52
10x-35-3=52
10x-38=52
10x=52+38
10x=90
x=9

Hope this helps :D
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y=-1.7+11y
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What is the domain of the function f(x) = x + 1/ x^2 - 6 + 8?
dangina [55]

Answer:

The domain is all values but x=4 and x=2

Step-by-step explanation:

f(x) = (x+1) / ( x^2-6x+8)

Factor the function

f(x) = (x+1) / ( (x-4) ( x-2))

The domain of the function is all values of x except where the function does not exist

This is where the denominator goes to zero

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Using the zero product property

x-4 =0   x-2 =0

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The height of a triangle is 5 cm shorter than its base. If the area of the triangle is 25 cm2, find the height of the triangle.
Olin [163]

Answer:

h = 5\ cm

Step-by-step explanation:

Let's call B at the base of the triangle and call h at the height of the triangle. Then we know that:

The height of a triangle is 5 cm shorter than its base. This means that:

h = B-5.

 The area of the triangle is 25 cm²

By definition the area of a triangle is:

A = 0.5Bh

For this triangle we know that A = 25\ cm^2 and h = B-5. We substitute these values in the equation and solve for B.

25 = 0.5B (B-5)

0.5B ^ 2-\frac{5}{2}B-25 = 0

Now we use the quadratic formula to solve the equation.

For an equation of the form ax ^ 2 + bx + c = 0 the quadratic formula is:

B=\frac{-b\±\sqrt{b^2-4ac}}{2a}

In this case note that:  a=0.5,\ \ b=-\frac{5}{2}\ \ c=-25

Then:

B=\frac{-(-\frac{5}{2})\±\sqrt{(-\frac{5}{2})^2-4(0.5)(-25)}}{2(0.5)}

B=\frac{\frac{5}{2}\±\sqrt{\frac{25}{4}+50}}{1}

B=\frac{5}{2}\±\frac{15}{2}

The solutions are:

B_1=\frac{5}{2}+\frac{15}{2}=10

B_2=\frac{5}{2}-\frac{15}{2}=-5

For this problem we take the positive solution.

B=10\ cm

Now we substitute the value of B in the equation to find the height h

h = 10-5

h = 5\ cm

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3 years ago
IF THREE DIAGONALS ARE DRAWN INSIDE A HEXAGON WITH EACH ONE PASSING THROUGH THE CENTER POINT OF THE HEXAGON, HOW MANY TRIANGLES
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Answer: 6 triangles

Step-by-step explanation:

hope this helps :)

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