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1 year ago

What is the slope intercept form for x+3y=0

Mathematics

1 answer:

Norma-Jean [14]1 year ago

3 0

Answer:

y = -1/3 x

Step-by-step explanation:

what is the slope intercept form for x+3y=0

Slope intercept form solves for y:

x + 3y = 0

subtract x from both sides:

x + 3y - x = 0 - x

3y = -x

divide both sides by 3:

3y/3 = -x/3

y = -1/3 x

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Find all points on the x-axis that are 14 units from the point (6,-7) All points on the x-axis that are 14 units from the point

Maksim231197 [3]

Answer: $(6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0)$ are the required points.

or (18.124,0) and ( -6.124,0) are the required points.

Step-by-step explanation:

Let (x,0) be the point on x -axis that are 14 units from the point (6,-7) .

Then by distance formula , we have

$\sqrt{(x-6)^2+(0-(-7))^2}=14\ \ \ [\ \because distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}]$

Taking square on both the sides , we get

$(x-6)^2+7^2=14^2\\\\\Rightarrow\ x^2+6^2-2(6)x+49=196\\\\\Rightarrow\ x^2+36-12x=147\\\\\Rightarrow\ x^2-12x=111\\\\\Rightarrow\ x^2-12x-111=0$

Using quadratic formula : $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\dfrac{12\pm\sqrt{(-12)^2-4(1)(-111)}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{144+444}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{588}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{2^2\times7^2\times3}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm14\sqrt{3}}{2}\\\\\Rightarrow\ x=6\pm7\sqrt{3}$

so, $(6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0)$ are the required points.

since $\sqrt{3}=1.732$

so, $(6+7(1.732),0)\text{ and }(6-7(1.732),0)$ are the required points.

i.e. (18.124,0) and ( -6.124,0) are the required points.

3 0

3 years ago

Please solve and explain the question below

solong [7]

Answer:

OPTION A

Step-by-step explanation:

To find the table substitute the points on the given function and compare the values.

The given function is: $y = x^2 + 2$ .

OPTION A:

(i) When x = -2

LHS = y = 6.

RHS = (-2)² + 2 = 4 + 2 = 6.

LHS = RHS

(ii) When x = -1

LHS = y = 3

RHS = (-1)² + 2 = 1 + 2 = 3.

LHS = RHS

(iii) When x = 0

LHS = y = 2

RHS = 0² + 2 = 2.

LHS = RHS

(iv) When x = 1

LHS = y = 3

RHS = (1)² + 2 = 1 + 2 = 3.

LHS = RHS

(v) When x = 2

LHS = y = 6

RHS = (2)² + 2 = 4 + 2 = 6

LHS = RHS

OPTION B:

(i) When x = -2

LHS = y = -2

RHS = (-2)² + 2 = 6

LHS $\neq$ RHS

OPTION B is eliminated.

OPTION C:

(i) When x = -2

Using the same reason as OPTION B this option is eliminated as well.

So, OPTION A is the correct answer.

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jenyasd209 [6]

Answer:

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Step-by-step explanation:

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crimeas [40]

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