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exis [7]
3 years ago
13

t{12" alt="2\sqrt{3 (\sqrt{27-2\sqrt{12" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Strike441 [17]3 years ago
8 0

Answer: =2\sqrt{3}\sqrt[4]{27-4\sqrt{3}}\quad \left(\mathrm{Decimal:\quad }\:7.33224\dots \right)

Step-by-step explanation:

2\sqrt{3\left(\sqrt{27-2\sqrt{12}}\right)}

=2\sqrt{3\sqrt{27-2\sqrt{12}}}

=2\sqrt{3}\sqrt[4]{27-4\sqrt{3}}

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3 (2x+3y)
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2cot^2x-3csc=0 find all solutions of the equation in the interval [0,2pi)
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(HELP) The point (- 8, - 2) is on the parent graph below. Where will it be on the transformed graph?
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Answer:

   x=\frac{y2 }{9}

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4 years ago
What is the solution of this system of linear equations? 3y =3/2 x + 6 1/2y – 1/4x = 3
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4 years ago
Write the equation of the line perpendicular to y=5/3x+2 that passes through the point (-3,5)
Makovka662 [10]

Answer

The equation of the line in point-slope form is

y - 5 = (-3/5) (x + 3)

We can simplify this to obtain the slope-intercept form

y - 5 = (-3/5) (x + 3)

y - 5 = (-3x/5) - (9/5)

y = (-3x/5) - (9/5) + 5

y = (-3x/5) + (16/5)

Then, to put the whole thing in a simplified equation form, we multiply through by 5

y = (-3x/5) + (16/5)

5y = -3x + 16

3x + 5y = 16

Explanation

The general form of the equation in point-slope form is

y - y₁ = m (x - x₁)

where

y = y-coordinate of a point on the line.

y₁ = This refers to the y-coordinate of a given point on the line

m = slope of the line.

x = x-coordinate of the point on the line whose y-coordinate is y.

x₁ = x-coordinate of the given point on the line

So, we just need to compute the slope of this line since a point on the line has already been provided (-3, 5)

For two lines with slopes m₁ and m₂ that are perpendicular to each other, the slopes are related through the relation

m₁m₂ = -1

And the slope-intercept form of the equation of a straight line is given as

y = mx + b

where

y = y-coordinate of a point on the line.

m = slope of the line.

x = x-coordinate of the point on the line whose y-coordinate is y.

b = y-intercept of the line.

So, the slope of the line whose equation is given in the question is

m₁ = (5/3)

This is obtained by comparing y = mx + b with y = (5/3)x + 2

So, we can obtain the slope of the line that we need

m₁m₂ = -1

m₁ = (5/3)

m₂ = ?

m₁m₂ = -1

(5/3) × m₂ = -1

(5m₂/3) = -1

Cross multiply

m₂ = (-3/5)

So, we can write the equation in point-slope form now

Recall that

y - y₁ = m (x - x₁)

m = slope = (-3/5)

Point = (x₁, y₁) = (-3, 5)

x₁ = -3

y₁ = 5

y - y₁ = m (x - x₁)

y - 5 = (-3/5) (x - (-3))

y - 5 = (-3/5) (x + 3)

Hope this Helps!!!

4 0
2 years ago
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