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Nitella [24]
3 years ago
8

Prime factorization of 180 math

Mathematics
2 answers:
RUDIKE [14]3 years ago
6 0

Answer:

5×2²×3²

Step-by-step explanation:

jenyasd209 [6]3 years ago
5 0
180|2
90|2
45|3
15|3
5|5
1

180=2^2\cdot3^2\cdot5
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2 years ago
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Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth.
likoan [24]

Answer:

Thus, the two root of the given quadratic equation x^2+4=6x is 5.24 and 0.76 .

Step-by-step explanation:

Consider, the given Quadratic equation, x^2+4=6x

This can be written as ,  x^2-6x+4=0

We have to solve using quadratic formula,

For a given quadratic equation ax^2+bx+c=0 we can find roots using,

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}  ...........(1)

Where,  \sqrt{b^2-4ac} is the discriminant.

Here, a = 1 , b = -6 , c = 4

Substitute in (1) , we get,

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

\Rightarrow x=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot 1 \cdot (4)}}{2 \cdot 1}

\Rightarrow x=\frac{6\pm\sqrt{20}}{2}

\Rightarrow x=\frac{6\pm 2\sqrt{5}}{2}

\Rightarrow x={3\pm \sqrt{5}}

\Rightarrow x_1={3+\sqrt{5}} and \Rightarrow x_2={3-\sqrt{5}}

We know \sqrt{5}=2.23607(approx)

Substitute, we get,

\Rightarrow x_1={3+2.23607}(approx) and \Rightarrow x_2={3-2.23607}(approx)

\Rightarrow x_1={5.23607}(approx) and \Rightarrow x_2=0.76393}(approx)

Thus, the two root of the given quadratic equation x^2+4=6x is 5.24 and 0.76 .

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A and B are complementary angles. If A = (x – 23)° and m
Lana71 [14]

Answer:

∠ B = 85°

Step-by-step explanation:

Complementary angles sum to 90° , then

x - 23 + 2x + 29 = 90 , that is

3x + 6 = 90 ( subtract 6 from both sides )

3x = 84 ( divide both sides by 3 )

x = 28

Then

∠ B = 2x + 29 = 2(28) + 29 = 56 + 29 = 85°

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2 years ago
write a slope intercept equation for a line passing through the point (2,-2) that is parallel and perpendicular to the line x=-1
vekshin1

You would need two different lines to complete this as lines cannot be both parallel and perpendicular (these are opposites). The answers would be:

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Perpendicular: y = -2

In order to find these, we first need to see that the original line of x = -1 is a horizontal line. Therefore, any line that is parallel should be horizontal as well. To get a horizontal line through the point (2, -2), the only option is x = 2.

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