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Brut [27]
2 years ago
5

The difference of two number is 3.Their sum is 13.Find the numbers

Mathematics
2 answers:
maw [93]2 years ago
5 0

Answer: 8 and 5

Step-by-step explanation:

x - y = 3

x + y = 13

------------------ (add the two equations)

2x = 16

x = 8

8 + y = 13 (substitute x as 8 into one of the equations)

y = 5

saul85 [17]2 years ago
3 0

Answer:

8,5

Step-by-step explanation:

x-y=3

x+y=13

2x=16

x=8, y=5

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Answer:

Simplify the equation as closely as possible to the form of y = mx + b. Check to see if your equation has exponents. If it has exponents, it is nonlinear (exponential). If your equation has no exponents, it is linear.

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2 years ago
A gas stove that normally sells for $749 is on sale at a 30% discount. What is the sale price of the gas stove
Virty [35]
The answer is 224.7 you got to turn 30% into decimal which is 0.30 
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6 0
3 years ago
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Divide: <br> -20.48<br> -----------<br> -4
marysya [2.9K]
If the question was -20.48 divided by -4 then the answer is 5.12
3 0
3 years ago
Round the number 14.63157 to the nearest ten-thousandth.
Anit [1.1K]

Answer:

14.6316

Step-by-step explanation:

The place value ten-thousandth falls on number 5. The following number is 7 and because 7 is greater than 5, "5" turns into 6.

Hope it helps!

8 0
3 years ago
Read 2 more answers
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
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