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3 years ago

What Is 4x+6y=48 and what is the general form and x intercept and y intercept and slope and factored form?

Mathematics

1 answer:

Vladimir [108]3 years ago

4 0

Given equation : 4x+6y=48 .

The general form of the linear equation is ax +by =c.

The given equation is already in general form.

Let us find x-intercept.

Plugging y =0 in given equation.

4x+6(0)=48 .

4x +0 = 48

4x = 48

Dividing both sides by 4, we get

4x/4 = 48/4

x = 12.

<h3>Therefore, x-intercept is (12,0).</h3>

Let us find y-intercept.

Plugging x =0 in given equation.

4(0)+6y=48 .

6y = 48

Dividing both sides by 6, we get

6y/6 = 48/6

y = 8.

<h3>Therefore, y-intercept is (0,8).</h3>

Let us convert given equation in slope-intercept form y=mx+b.

4x+6y=48

Subtracting 4x from both sides, we get

4x-4x+6y=48-4x

6y = -4x+48

Dividing both sides by 6, we get

6y/6 = -4x/6+48/6

y = -2/3 x +8.

<h3>Therefore, slope is -2/3.</h3>

Let us find factored form now.

GCF of the given linear equation is 2.

Factoring out 2 in all term, we get

3(2x+3y=24)

That is factored form.

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3 years ago

last month maria hiked a 5 mile trail several times and a 10 mile trail several times let x represent the number of times she hi

Vlada [557]

Answer:

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Step-by-step explanation:

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2 years ago

the slope of a line passing through h (-2, 5) is -3/4. Which ordered pair represents a point on this line?

rosijanka [135]

Answer:

Step-by-step explanation:

Calculate the slope of the given points with the point (- 2, 5 )

If the slope is - $\frac{3}{4}$ then the point is on the line

Calculate slope m using the slope formula

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ ) = (- 2, 5) and (x₂, y₂ ) = (6, - 1)

m = $\frac{-1-5}{6+2}$ = $\frac{-6}{8}$ = - $\frac{3}{4}$ ← point (6, - 1) is on the line

Repeat with (x₁, y₁ ) = (- 2, 5 ) and (x₂, y₂ ) = (2, 8)

m = $\frac{8-5}{2+2}$ = $\frac{3}{4}$ ← point (2, 8) is not on the line

Repeat with (x₁, y₁ ) = (- 2, 5) and (x₂, y₂ ) = (- 5, 1)

m = $\frac{1-5}{-5+2}$ = $\frac{-4}{-3}$ = $\frac{4}{3}$ ← point (- 5, 1) is not on the line

Repeat with (x₁, y₁ ) = (- 2, 5) and (x₂, y₂ ) = (1, 1)

m = $\frac{1-5}{1+2}$ = - $\frac{4}{3}$ ← point (1, 1) is not on the line

Thus the point on the line is (6, - 1 ) → A

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