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n200080 [17]
2 years ago
6

Find the equation of the line Which passes through ( -4,-2) and it is parallel to 2y-6x-3 =0

Mathematics
1 answer:
lora16 [44]2 years ago
6 0

Answer:

y = 3x + 10

Step-by-step explanation:

In this question, we have to find an equation of a line that's parallel to the equation given.

First, lets turn the given equation into slope-intercept form:

2y - 6x - 3 = 0

Add 6x and 3 to both sides.

2y = 6x + 3

Divide both sides by 2.

y = 3x + 3/2

We know that the slope of our parallel line has to be 3, now we just need to find what our beginning point, or y-intercept, is.

To find our parallel line, plug in the coordinates to the y = 3x + b and solve.

Solve for b:

-2 = 3(-4) + b

-2 = -12 + b

Add 12 to both sides.

10 = b

Our parallel equation would be: y = 3x + 10

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Answer:

x = 70°

Step-by-step explanation:

x = one of the congruent angles in the triangle

Given angle measure: 40°

x = ? + ? = 180 - 40 = 140

x = 70

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Because vol= length x width x height , and assuming that the height is 11.5 because breadth means width but you already have a width then, vol= 11.5 x 7.4 x 11.5, vol = 978.65
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X+10=1/3(5x+10)<br><br> How do you solve this equation?
Nikitich [7]

Answer:

x = 10

Step-by-step explanation:

First, subtract 10 from both sides

Next, Simplify

Then, subtract 5/3x from both sides

After, simplify


Lastly, multiply both sides by 3


Then finally simplify for your final answer of x = 10


Hope this helps! :)

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2 years ago
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Which of the following are identities? Check all that apply
Natasha2012 [34]

Answer:

A, C

Step-by-step explanation:

Actually, those questions require us to develop those equations to derive into trigonometrical equations so that we can unveil them or not. Doing it only two alternatives, the other ones will not result in Trigonometrical Identities.

Examining

A) True

\frac{1-tan^{2}x}{2tanx} =\frac{1}{tan2x} \\ \frac{1-tan^{2}x}{2tanx} =\frac{1}{\frac{2tanx}{1-tan^{2}x}}\\ tan2x=\frac{1-tan^{2}x}{2tanx}

Double angle tan2\alpha =\frac{1 -tan^{2}\alpha }{2tan\alpha}

B) False,

No further development towards a Trig Identity

C) True

Double Angle Sine Formula sin2\alpha =2sin\alpha *cos\alpha

sin(8x)=2sin(4x)cos(4x)\\2sin(4x)cos(4x)=2sin(4x)cos(4x)

D) False No further development towards a Trig Identity

[sin(x)-cos(x)]^{2} =1+sin(2x)\\ sin^{2} (x)-2sin(x)cos(x)+cos^{2}x=1+2sinxcosx\\ \\sin^{2} (x)+cos^{2}x=1+4sin(x)cos(x)

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Marina86 [1]

There are infinitely many ways to do this. One such way is to draw a very thin stretched out rectangle (say one that is very tall) and a square. Example: the rectangle is 100 by 2, while the square is 4 by 4.

Both the rectangle and the square have the same corresponding angle measures. All angles are 90 degrees.

However, the figures are not similar. You cannot scale the rectangle to have it line up with the square. The proportions of the sides do not lead to the same ratio

100/4 = 25

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so 100/4 = 2/4 is not a true equation. This numerically proves the figures are not similar.

side note: if you are working with triangles, then all you need are two pairs of congruent corresponding angles. If you have more than three sides for the polygon, then you'll need to confirm the sides are in proportion along with the angles being congruent as well.

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