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gayaneshka [121]
2 years ago
14

The equation represents the area, A, of a rectangle, in square units. A=x² - x - 6

Mathematics
1 answer:
vlada-n [284]2 years ago
4 0

Answer:

ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh hhhhhhhhhhhhhhhh\hhhh

Step-by-step explanation:

nolos

e  

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SOMEONE PLEASE HELP ME ASAP PLEASEEEE!!!!!!​
Flauer [41]

Answer:

i maybe wrong

Step-by-step explanation:

I think you take the formula (x+3, y-4) and u replace it with the P and Q points

so (2+3)=5

and (4-4)=0

and the same with the Q point

so its -3 + 3 = 0

and 2-4 = -2

so now u have

P' = ( 5,0) and Q'= (0,-2)

now the slope for that is also 2/5

6 0
3 years ago
What property of an even function do you see in this graph?
Brut [27]

Answer:

Step-by-step explanation:

I'm confused about what you want me to do but I know for a faction it is a function.

6 0
3 years ago
WILL GIVE BRAINLIEST. Point C is at (-3, -2). After being reflected across the ​ x-axis, what are the coordinates of Point C?​ (
ZanzabumX [31]

Answer:

the answer is (3, 2)

Step-by-step explanation:

6 0
3 years ago
Which point lies on the circle represented by the equation (x+5) + (y-9) = 8
olga_2 [115]

Answer:

(-5,9+√8)

Step-by-step explanation:

There are two ways in which you can find a point that lies in the circle. One of them is to do y the subject of the formula, and another one is to determine the center of the circumference, and with the information of the radius, you can sum this value upward or downward.

the general equation of a circle is:

(x-h)^2+(y-k)^2=r^2

with center at (h,k)

you have the following equation:

(x+5)^2+(y-9)^2=8\\\\h=-5\\\\k=9\\\\r=\sqrt{8}

Then, the center is (-5,9)

if you sum the value of the radius in one of the fourth directions (up, down, left, right), for example upward you have

(-5,9+r)=(-5,9+\sqrt{8})

Then, one point that lies in the circle is (-5,9+√8)

3 0
3 years ago
Multiply two and five-eighths negative two and three-fifths .
Yuki888 [10]
The answer is -6.825 in fraction form, it would be -6 33/40
4 0
3 years ago
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