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4 years ago

17. Find an equation for the line passing

Mathematics

1 answer:

Ksivusya [100]4 years ago

5 0

Answer:

Step-by-step explanation:

(3,2),(6,3)

slope = (y2 - y1) /(x2 - x1)

slope = (3 - 2) / (6 - 3) = 1/3

y = mx + b

slope(m) = 1/3

use either of ur points...(3,2)...x = 3 and y= 2

now sub and find b, the y int

2 = 1/3(3) + b

2 = 1 + b

2 - 1 = b

1 = b

so ur equation is : y = 1/3x + 1

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3 years ago

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3 years ago

What are the verter and x-intercepts of the graph of the function given below?<br> y=x2-2x-35

horsena [70]

For x intercepts, plug in 0 for y.

0 = (x^2) - 2x - 35

*factoring* = (x-7)(x+5)

x intercepts = 7,-5

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so,

x = -b/2a = -(-2)/2 = 1

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3 years ago

If tanA=a <br>then find sin4A-2sin2A/ sin4A+2sin2A

anygoal [31]

Answer:

The value of the given expression is

$\frac{sin4A-2sin2A}{sin4A+2sin2A}=-a^2$

Step by step Explanation:

Given that $tanA=a$

To find the value of $\frac{sin4A-2sin2A}{sin4A+2sin2A}$

Let us find the value of the expression :

$\frac{sin4A-2sin2A}{sin4A+2sin2A}=\frac{2cos2Asin2A-2sin2A}{2cos2Asin2A+2sin2A}$ ( by using the formula $sin2A=2cosAsinA$ here A=2A)

$=\frac{2sin2A(cos2A-1)}{2sin2A(cos2A+1)}$

$=\frac{(cos2A-1)}{(cos2A+1)}$

$=\frac{(-(1-cos2A))}{(1+cos2A)}$ (using $sin^2A+cos^2A=1$ here A=2A)

$=\frac{-(sin^2A+cos^2A-(cos^2A-sin^2A))}{sin^2A+cos^2A+(cos^2A-sin^2A)}$ (using $cos2A=cos^2A-sin^2A$ here A=2A)

$=\frac{-(sin^2A+cos^2A-cos^2A+sin^2A)}{sin^2A+cos^2A+(cos^2A-sin^2A)}$

$=\frac{-(sin^2A+sin^2A)}{cos^2A+cos^2A}$

$=\frac{-2sin^2A}{2cos^2A}$

$=-\frac{sin^2A}{cos^2A}$

$=-tan^2A$ ( using $tanA=\frac{sinA}{cosA}$ here A=2A )

$=-a^2$ (since tanA=a given )

Therefore $\frac{sin4A-2sin2A}{sin4A+2sin2A}=-a^2$

6 0

3 years ago