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just olya [345]
3 years ago
5

question is in picture. Plzzz help me. Due in 2hrs

Mathematics
1 answer:
Bad White [126]3 years ago
4 0

Answer:

I think it's line "A"

Step-by-step explanation:

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What is the y-coordinate of point D after a translation of (x, y) ? (x + 6, y – 4)?
Mandarinka [93]

Answer:

Y-coordinate = y - 4

Step-by-step explanation:

In the given question point D coordinate is (x, y)

and D after translation

x goes to 6 unit to the right and y goes to 4 unit down.

Hence,

          Y-coordinate = y-4

Example:From graph

              Let point D coordinate (x , y) = (2 , 3)

After rotation

              D coordinates is (x+6 , y-4) = (2+6 , 3_4) = (8 , -1)


6 0
3 years ago
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What is the measure of the angle? A 125, B135, C 55, or D 45.
Amiraneli [1.4K]
I think the answer is C, I could be wrong though
5 0
3 years ago
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1. Evaluate 4x' + 7xy - 5y for x = 5, and y = 3.​
Nataly_w [17]

Answer:

110 is your answer

7 0
3 years ago
Need help on finding the equation for both graphs
Assoli18 [71]

for #1 it is y=-3/2x+4 the -3/2 is the slope and the 4 is the y intercept

for #2 it is y=6/+2 the 6/2 is the slope and the 2 is the y-intercept

the y-intercept is where the line passes through the y axis

5 0
3 years ago
Use implicit differentiation to find an equation of the tangent line to the curve at the given points. (x2 + y2)2 = 3x2y − y3;
Andrew [12]

Answer:

a.y+1=0

b.2x+4y=1

Step-by-step explanation:

We are given that

(x^2+y^2)^2=3x^2y-y^3

a.(0,-1)

Differentiate w.r.t x

2(x^2+y^2)(2x+2yy')=6xy+3x^2y'-3y^2y'.....(1)

Substitute x=0 and y=-1 in equation (1)

2(0+1)(-2y')=-3y'

-4y'+3y'=0

-y'=0

y'=0

m=y'=0

Point-slope form:

y-y_0=m(x-x_0)

Using the formula

y+1=0

This is required equation of tangent line to the given curve at point (0,-1).

b.(-1/2,1/2)

Substitute the value in equation (1)

2(1/4+1/4)(-1+y')=6(-1/2)(1/2)+3(1/4)y'-3(1/4)y'

2(2/4)(-1+y')=-3/2+3/4y'-3/4y'

-1+y'=-3/2

y'=-3/2+1=\frac{-3+2}{2}=-\frac{1}{2}

m=y'=-1/2

Again using point-slope formula

y-1/2=-1/2(x+1/2)

\frac{2y-1}{2}=-\frac{1}{4}(2x+1)

2y-1=-\frac{1}{2}(2x+1)

4y-2=-2x-1

2x+4y=2-1

2x+4y=1

5 0
3 years ago
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