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3 years ago

Which of the following points lies on the x axies : (a) -6,1 (b) 0,7 (c) 8,0 (d) 5,10

Mathematics

1 answer:

fenix001 [56]3 years ago

7 0

Answer:

B 0,7

AS O IS THE MID POINT OF THE INTERSECTION 9F X AND Y AXIS SO IT LIES ON X AXIS

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3 years ago

Write an equation for the line that passes through the points (0, -6) and (-3, 0).

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$\displaystyle y = -2x -6$

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3 years ago

Solve y = x + 4 for x. x = y − 4 x = y + 4 x = −y + 4 x = −y − 4

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4 years ago

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Hey guys im new here please solve this for me with steps! ill mark as the best answer

Vinil7 [7]

Answer:

The factors of $2(x+y)^2-9(x+y)-5$ is ((x+y)-5)(2x+2y+1)

Step-by-step explanation:

Given polynomial

=> $2(x+y)^2-9(x+y)-5$

To Find:

The factors of the polynomial =?

Solution:

Lets assume k = (x+y)

Then $2(x+y)^2-9(x+y)-5$ can be written as $2k^2-9k-5$

Now by using quadratic formula

k = $\frac{-b\pm\sqrt{(b^2-4ac}}{2a}$

where

a= 2

b= -9

c= -5

Substituting the values, we get

k = $\frac{-b\pm\sqrt{(b^2-4ac)}}{2a}$

k = $\frac{-(-9) \pm \sqrt{((-9)^2-4(2)(-5)}}{2(2))}$

k = $\frac{-(-9) \pm \sqrt{(81+40)}}{4}$

k = $\frac{-(-9) \pm \sqrt{(121)}}{4}$

k = $\frac{-(-9) \pm 11}}{4}$

k= $\frac{ 9 \pm 11}{4}$

k = $\frac{20}{4}$ k = $\frac{-2}{4}$

$k_1 =5$ $k_2 = -\frac{1}{2}$

$2k^2-9k-5= 2(k-5)(k+\frac{1}{2})$

Solving the RHS we get

$\frac{2}{2}(k-5)(2k+1)$

$(k-5)(2k+1)$

Substituting k = x+y

$((x+y)-5)(2(x+y+1)$

$((x+y)-5)(2x+2y+1)$

5 0

3 years ago

Which of the following points lies on the x axies : (a) -6,1 (b) 0,7 (c) 8,0 (d) 5,10​

Which of the following points lies on the x axies : (a) -6,1 (b) 0,7 (c) 8,0 (d) 5,10