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3 years ago

How do you graph a line with one number y= -4

Mathematics

1 answer:

-BARSIC- [3]3 years ago

5 0

Answer:

See explanation

Step-by-step explanation:

When you have an equation like that, all you have to do is draw a line at the y-value given. In our case, y = -4 is a horizontal line. Graph paper is made of many intersecting lines. From the x-axis (should be a big black line that moves left to right), move down four lines. Once you get to the fourth line, trace the line from left to right. Put some points at the y-value of -4 (for example (-1, -4)) if it helps you straighten the line.

Some lines may be vertical and move up and down. Those lines have an x instead of a y (x=3, x= -12, etc.). Fortunately, these lines, both vertical and horizontal, are a lot easier to draw than other equations.

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If sinA=√3-1/2√2,then prove that cos2A=√3/2 prove that

Ivan

Answer:

$\boxed{\sf cos2A =\dfrac{\sqrt3}{2}}$

Step-by-step explanation:

Here we are given that the value of sinA is √3-1/2√2 , and we need to prove that the value of cos2A is √3/2 .

Given :-

• $\sf\implies sinA =\dfrac{\sqrt3-1}{2\sqrt2}$

To Prove :-

• $\sf\implies cos2A =\dfrac{\sqrt3}{2}$

Proof :-

We know that ,

$\sf\implies cos2A = 1 - 2sin^2A$

Therefore , here substituting the value of sinA , we have ,

$\sf\implies cos2A = 1 - 2\bigg( \dfrac{\sqrt3-1}{2\sqrt2}\bigg)^2$

Simplify the whole square ,

$\sf\implies cos2A = 1 -2\times \dfrac{ 3 +1-2\sqrt3}{8}$

Add the numbers in numerator ,

$\sf\implies cos2A = 1-2\times \dfrac{4-2\sqrt3}{8}$

Multiply it by 2 ,

$\sf\implies cos2A = 1 - \dfrac{ 4-2\sqrt3}{4}$

Take out 2 common from the numerator ,

$\sf\implies cos2A = 1-\dfrac{2(2-\sqrt3)}{4}$

Simplify ,

$\sf\implies cos2A = 1 -\dfrac{ 2-\sqrt3}{2}$

Subtract the numbers ,

$\sf\implies cos2A = \dfrac{ 2-2+\sqrt3}{2}$

Simplify,

$\sf\implies \boxed{\pink{\sf cos2A =\dfrac{\sqrt3}{2}} }$

Hence Proved !

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