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tangare [24]
3 years ago
9

Please help me!!! ASAPP

Mathematics
1 answer:
larisa86 [58]3 years ago
5 0

Answer:

x=2 y=8 ---> (2;8)

Step-by-step explanation:

2x+5y=44

6x-4=y

1.

2x+5(6x-4)=44

2x+30x-20=44

32x=64

x=2

2.

6x-4=y

6*2-4=y

12-4=y

y=8

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Please help me. i want to have this finished.
Anuta_ua [19.1K]
We know that y is equal to x-2. We can just substitute x-2 for y since y is equal to it. 

10x- 9(x-2)=24

Distribute.

10x- 9x+18= 24

x+18= 24

Subtract 18 on both sides.

x= 6

Now, plug in x.

y= (6)-2

y= 4

We can check this to see if this works:

4= 6-2, 4=4 

10(6)- 9(4)= 24

60-36= 24, 24=24

x=6 and y=4

I hope this helps!
<em>~kaikers</em>
5 0
3 years ago
Need some help please
spayn [35]
It would be A.) 7. hope this helps
3 0
3 years ago
Read 2 more answers
3 less than c in numbers
Ira Lisetskai [31]

Answer:

Step-by-step explanation:

3<3

7 0
2 years ago
Is the given point interior , exterior, or on the circle k (x+2)2 + (y-3)2 =18 P (8,4)
Mnenie [13.5K]
One way would be to find the distance from the point to the center of the circle and compare it to the radius

for
(x-h)^2+(y-k)^2=r^2
the center is (h,k) and the radius is r

and the distance formula is
distance between (x_1,y_1) and (x_2,y_2) is
D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}


r=radius
D=distance form (8,4) to center

if r>D, then (8,4) is inside the circle
if r=D, then (8,4) is on the circle
if r<D, then (8,4) is outside the circle


so
(x+2)^2+(y-3)^2=18
(x-(-2))^2+(y-3)^2=(\sqrt{18})^2
(x-(-2))^2+(y-3)^2=(3\sqrt{2})^2

the radius is 3\sqrt{2}
center is (-2,3)

find distance between (8,4) and (-2,3)

D=\sqrt{(8-(-2))^2+(4-3)^2}
D=\sqrt{(8+2)^2+(1)^2}
D=\sqrt{10^2+1}
D=\sqrt{100+1}
D=\sqrt{101}




r=3\sqrt{2}≈4.2
D=\sqrt{101}≈10.04

do r<D

(8,4) is outside the circle

6 0
3 years ago
(-5.975)(16.59) muiltilpy
olga nikolaevna [1]

Answer:

-99.12525

Step-by-step explanation:

4 0
3 years ago
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