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Ivanshal [37]
3 years ago
14

Solve the system using subtitions {5x-y=5 5x3y=15

Mathematics
1 answer:
denis23 [38]3 years ago
6 0

Answer:

I just googled it

Step-by-step explanation:

5 x − y = 15

Add  y  to both sides of the equation.

5 x = 15 + y

Divide each term by  5  and simplify.

x= 3 + y 5

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What is the equation for the translation of x2 + y2 = 16 seven units to the right and five units up?
asambeis [7]

Answer:

  • (x − 7)2 + (y − 5)2 = 16

Step-by-step explanation:

The given circle has equation

\sf \: x^2+y^2=16x

The equation of a circle with center (h,k) and radius r units is

\sf(x-h)^2+(y-k)^2=r^2(x−h)

\sf(x-7)^2+(y-5)^2=4^2(x−7)

\sf(x-7)^2+(y-5)^2=16(x−7)

<h2>❖ Tip❖ :- </h2>

This is the equation that has its center at the origin with radius 4 units.

When this circle is translated seven units to the right and five units up, then the center of the circle will now be at (7,5).

6 0
2 years ago
If ƒ(x ) = x^2 + 1 and g(x ) = 3x + 1, find [ƒ(2) - g(1)]^2.<br><br> 1<br> 2<br> 9
stepladder [879]
I think that the answer would be 2 if I am right 
5 0
3 years ago
Read 2 more answers
Is {(5, 12), (-4,9), (-2,-7), (-4,0), (3, 2)}<br> A function? Please help me
Anna71 [15]

Answer:

Not a function

Step-by-step explanation:

(-4,9), (-4,0)   : one input (-4) with 2 output -9 , 0

6 0
3 years ago
Find the valuesof integers x and y: x^(8)+16y^(8)=0
avanturin [10]

Answer:

x=0,

y=0

Step-by-step explanation:

x^8+16y^8=0\\(x^{2}) ^{4} +16(y^{2}) ^{4} =0\\x^2\geq 0,y^2\geq 0\\sum~ of ~two ~positive~ integers ~is~ zero~ if~ each=0\\so~x=0,~y=0

8 0
4 years ago
Find the equation of locus of a point which moves such that its distance from (0,2) is one third distance from (-2,3). ( I WILL
scoray [572]

Answer:

8(x^2+y^2)-4x-30y+23=0

Step-by-step explanation:

<u />

<u>Distance formula</u>

\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Let P(x, y) = any point on the locus

Let A = (0, 2)    

Let B = (-2, 3)

If a point moves such that its distance from (0, 2) is one third distance from (-2, 3):

PA=\dfrac{1}{3}PB

Therefore, using the distance formula:

\implies \sqrt{(x-0)^2+(y-2)^2}=\dfrac{1}{3}\sqrt{(x-(-2))^2+(y-3)^2}

Square both sides:

\implies x^2+(y-2)^2=\dfrac{1}{9}[(x+2)^2+(y-3)^2]

\implies x^2+y^2-4y+4=\dfrac{1}{9}(x^2+4x+4+y^2-6y+9)

Multiply both sides by 9:

\implies 9x^2+9y^2-36y+36=x^2+4x+4+y^2-6y+9

\implies 8x^2+8y^2-4x-30y+23=0

\implies 8(x^2+y^2)-4x-30y+23=0

3 0
2 years ago
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