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2 years ago

What are the Horizontal and vertical shift of y = x+2-3

Mathematics

1 answer:

Nana76 [90]2 years ago

7 0

If the equation is y=(x+2)-3 then the shifts are
3 down and 2 to the left

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PLEASE HELP<br><br>divide. (3x-2)(x-4)-(x-4)(6-5x)<br>/(4-x)(8x-1)

8_murik_8 [283]

Answer:

x = 39/74 - sqrt(1817)/74 or x = 39/74 + sqrt(1817)/74

Step-by-step explanation:

Solve for x:

(x - 4) (3 x - 2) - ((6 - 5 x) (x - 4) (8 x - 1))/(4 - x) = 0

Simplify and substitute y = 4 - x.

(x - 4) (3 x - 2) - ((6 - 5 x) (x - 4) (8 x - 1))/(4 - x) = -434 + 257 (4 - x) - 37 (4 - x)^2

= -37 y^2 + 257 y - 434:

-37 y^2 + 257 y - 434 = 0

Divide both sides by -37:

y^2 - (257 y)/37 + 434/37 = 0

Subtract 434/37 from both sides:

y^2 - (257 y)/37 = -434/37

Add 66049/5476 to both sides:

y^2 - (257 y)/37 + 66049/5476 = 1817/5476

Write the left hand side as a square:

(y - 257/74)^2 = 1817/5476

Take the square root of both sides:

y - 257/74 = sqrt(1817)/74 or y - 257/74 = -sqrt(1817)/74

Add 257/74 to both sides:

y = 257/74 + sqrt(1817)/74 or y - 257/74 = -sqrt(1817)/74

Substitute back for y = 4 - x:

4 - x = 257/74 + sqrt(1817)/74 or y - 257/74 = -sqrt(1817)/74

Subtract 4 from both sides:

-x = sqrt(1817)/74 - 39/74 or y - 257/74 = -sqrt(1817)/74

Multiply both sides by -1:

x = 39/74 - sqrt(1817)/74 or y - 257/74 = -sqrt(1817)/74

Add 257/74 to both sides:

x = 39/74 - sqrt(1817)/74 or y = 257/74 - sqrt(1817)/74

Substitute back for y = 4 - x:

x = 39/74 - sqrt(1817)/74 or 4 - x = 257/74 - sqrt(1817)/74

Subtract 4 from both sides:

x = 39/74 - sqrt(1817)/74 or -x = -39/74 - sqrt(1817)/74

Multiply both sides by -1:

Answer: x = 39/74 - sqrt(1817)/74 or x = 39/74 + sqrt(1817)/74

7 0

3 years ago

If x = 9, what is the value of 6(x - 2)?

balandron [24]

Answer: 42

6(9-2)

= 6 * 7

= 42

3 0

2 years ago

Read 2 more answers

What are the first three terms of a geometric sequence in which a5= 25 and the common ratio is 5?

shepuryov [24]

Answer:

1/25, 1/5, 1

Step-by-step explanation:

8 0

2 years ago

Simplify <br> (3-3)/√3=0

zalisa [80]

Answer:

its 0 anyways lol

Step-by-step explanation:

4 0

3 years ago

there has to be at least one operating path. Once a component fails on a path, that path is no longer operating. The reliability

Eva8 [605]

Answer:

$R = P(R_1) P(R_2|R_1) P(R_3|R_2)*....*P(R_n |R_1,..., R_{n-1}$

For this case the reliability of the sytem would be given by:

$R= \prod_{i=1}^n R_i$

R1= 0.95, R2 =0.95, R3= 0.5, R4 = 0.79 , R5 = 0.6

And replacing we got:

$R = 0.95*0.95*0.5*0.79*0.6= 0.21389$

Step-by-step explanation:

We can assume that the system work in series

If we have in general n units the reliability of the system is the probability that unit 1 succeeds and unit 2 succeeds and all of the other units in the system succeed. fror n units must succeed for the system to succeed. The reliability of the system is then given by:

$R = P(R_1) P(R_2|R_1) P(R_3|R_2)*....*P(R_n |R_1,..., R_{n-1}$

For this case the reliability of the sytem would be given by:

$R= \prod_{i=1}^n R_i$

R1= 0.95, R2 =0.95, R3= 0.5, R4 = 0.79 , R5 = 0.6

And replacing we got:

$R = 0.95*0.95*0.5*0.79*0.6= 0.21389$

8 0

3 years ago