All categories

2 years ago

Help plsssss????!??,????

Mathematics

2 answers:

NeX [460]2 years ago

7 0

Answer:

i'll post this answer since the question was already answered, thank you so much for the points! :) <3

Strike441 [17]2 years ago

3 0

Answer:

y=4x+5

Step-by-step explanation:

so like the b in y= mx +b is the starting point well in that graph it is also is where the line touches the y-axis and the slope is rise/run so start at 5 move over 1 then go up 4 that 4/1 = 4 so m=4 so if y= 4x+b and b being 5 then y=4x+5

You might be interested in

Jessica works between 12h and 18h per week at her part-time job. She earns $9.25 per hour. The function p(h) = 9.25h models the

Tpy6a [65]

D. All real numbers between 111 and 166.5 inclusive.

3 0

3 years ago

Find the product -1.2(0.04)

patriot [66]

Answer:

-0.048

Step-by-step explanation:

-1.2 * 0.04 = -1.2 * 4 / 100 = -4.8 / 100 = -).048

4 0

2 years ago

larisa86 [58]

Answer:

The correct answer is D, 0.20. You add the probabilities of either situation occurring and subtract the probability of both of them occurring.

Step-by-step explanation:

0.25 + 0.08 - 0.13 = 0.20

3 0

2 years ago

Read 2 more answers

39. The diagram shows Luciana and Tanner's location

AlexFokin [52]

Answer:

2<d<8

Step-by-step explanation:

I have the same packet

4 0

2 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

2 years ago