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3241004551 [841]
3 years ago
15

What is the square root of 69

Mathematics
2 answers:
Alex73 [517]3 years ago
8 0
It is 8.306623863 so about 8.3
Rasek [7]3 years ago
6 0
The square root of 69 is 8.30.because 8.30*8.30 is 69
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The length of a rectangle is the sum of the width and one. The area direct angle 72 units. What’s the length, in units, of the r
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Answer:

The length of the rectangle is of 9 units.

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

Area of a rectangle:

A rectangle has width w and length l. The area is the multiplication of these measures, that is:

A = wl

The length of a rectangle is the sum of the width and one.

This means that l = w+1, or w = l - 1

The area direct angle 72 units. What’s the length, in units, of the rectangle

We want to find the length. So

wl = 72

(l-1)l = 72

l^2 - l - 72 = 0

Quadratic equation with a = 1, b = -1, c = -72. So

\bigtriangleup = (-1)^{2} - 4(1)(-72) = 289

l_{1} = \frac{-(-1) + \sqrt{289}}{2*(1)} = 9

l_{2} = \frac{-(-1) - \sqrt{289}}{2*(1)} = -8

Since the length is a positive measure, the length of the rectangle is of 9 units.

8 0
2 years ago
Write the equation of a line that is perpendicular to y=-x-6y=−x−6y, equals, minus, x, minus, 6 and that passes through the poin
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The given line has a slope of -1 so the perpendicular line will have a slope of -1/1=1
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The formula of a midpoint:

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Substitute

\dfrac{5+x_B}{2}=1.5\qquad|\cdot2\\\\5+x_B=3\qquad|-5\\\\x_B=-2\\\\\dfrac{7+y_B}{2}=1.5\qquad|\cdot2\\\\7+y_B=3\qquad|-7\\\\y_B=-4

<h3>Answer: (-2, -4)</h3>
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3 years ago
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stiks02 [169]

Answer:

C) -7/3

Step-by-step explanation:

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simplify -14/6 to -7/3

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Four subtracted from the reciprocal of a number
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four subtracted from this
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