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2 years ago

Need heeeelp please

Mathematics

1 answer:

Fiesta28 [93]2 years ago

5 0

Answer:

4 $v^{7}$ x³(7 $y^{6}$ - 3v² $x^{6}$ )

Step-by-step explanation:

28 $v^{7}$ x³ $y^{6}$ - 12 $v^{9}$ $x^{9}$ ← factor out 4 $v^{7}$ x³ from each term

= 4 $v^{7}$ x³(7 $y^{6}$ - 3v² $x^{6}$ )

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Which of the following points would be on the graph of the equation y=−4x+6?(−4, −10)(−4, 22)(1, −10)(1, −2)

suter [353]

Answer:

Only the point (-4, 22) is ON THE GRAPH of the equation. Please also check the attached graph.

Step-by-step explanation:

Given the equation

$y=-4x+6$

Let us substitute all the given points to check which points satisfy the equation.

Substitute (-4, -10) in the equation

$y=-4x+6$

$-10=-4(-4)+6$

$-10=16+6$

$-10=22$

As the R.H.S ≠ L.H.S, hence (-4, -10) does not satisfy the equation.

Hence, (-4, -10) is NOT ON THE GRAPH of the equation

Substitute (-4, 22) in the equation

$y=-4x+6$

$22 = -4(-4)+6$

$22 = 16+6$

$22 = 22$

As the R.H.S = L.H.S, hence (-4, 22) satisfies the equation.

Hence, (-4, 22) is ON THE GRAPH of the equation. Please also check the attached graph.

Substitute (1, -10) in the equation

$y=-4x+6$

$-10=-4(1)+6$

$-10=-4+6$

$-10=2$

As the R.H.S ≠ L.H.S, hence (1, -10) does not satisfy the equation.

Hence, (1, -10) is NOT ON THE GRAPH of the equation

Substitute (1, -2) in the equation

$y=-4x+6$

-2 = -4(1)+6

-2 = -4 + 6

-2 = 2

As the R.H.S ≠ L.H.S, hence (1, -2) does not satisfy the equation.

Hence, (1, -2) is NOT ON THE GRAPH of the equation

Therefore, from the above analysis, we conclude that only the point (-4, 22) is ON THE GRAPH of the equation. Please also check the attached graph.

8 0

3 years ago

Need heeeelp please ​

Need heeeelp please