Write the equation of the line that passes through (4, 2) and is parallel to the line y = 2x – 1.

Answers

Answer 1

Answer: y= 2x-6
for the line to be parallel with y=2x-1 it has to have the same gradient which is the 2x part. and to get it to pass through (4,2) it has to be -6

Answer 2

Answer:

Answer:

Step-by-step explanation:

2x – y = 6

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How to find the area of a right triangle?

Answers

A=Ab/2 is how I find it

Write as a product: a-b+a^2-B^2

Answers

Answer: (a - b)(a + b + 1)

a - b + a² - b²

= (a - b) + (a - b)(a + b)

= (a - b)(a + b + 1)

Step-by-step explanation:

Find the next three terms in the arithmetic sequence: 25, 34, 43, 52, ...

Answers

Answer:

61,70,79

Step-by-step explanation:

common difference=9

It is an arithmetic progression.

therefore next 3 terms of the sequence is 61,70,79

Answer:

61, 70,79

Step-by-step explanation:

34 - 25 = 9

43 - 34 = 9

52 - 43 = 9

52 + 9 =61

the pattern is plus 9

Simplify and write your answer without exponents 16^(-5/4)

Answers

$16^{ -(5)/(4)} \n \n 2^(-5) \ / \ simplify \n \n (1)/(2^5) \ / \ negative \ power \ rule \n \n (1)/(32) \ / \ simplify \n \n Answer: \fbox {1/32} \ or \fbox {0.0313}$

Let f(x)=2x-3 and g(x)= -x^2-1 find ( g o f) (x)

Answers

$f(x)=2x-3\ \ \ and\ \ \ g(x)= -x^2-1\n\n( g \circ f) (x)=g\bigg(f(x)\bigg)=g\bigg(2x-3\bigg)=-(2x-3)^2-1=\n\n=-(4x^2-12x+9)-1=-4x^2+12x-9-1=-4x^2+12x-10$

$g \circ f=-(2x-3)^2-1=-(4x^2-12x+9)-1=\n =-4x^2+12x-9-1=-4x^2+12x-10\n$

5q+4 -11/2= 2q + 3/2(2q-1)

Answers

Answer:

Interval Notation:

(−∞,∞)

Step-by-step explanation:

Any value of q makes the equation true.

Hope this helps :)

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