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3 years ago

Sketch a graph of the linear relation 8x+2y = 24. Explain the procedure.

1 answer:

6 0

First we need the slope and the y int which can be found by putting ur equation in y = mx + b form, where m is ur slope and b is ur y int.

8x + 2y = 24
2y = -8x + 24
y = -4x + 12.....so the slope is -4, and the y int is 12 or (0,12)

to find the x int, sub in 0 for y and solve for x....in either the original equation or the slope intercept equation
8x + 2y = 24
8x + 2(0) = 24
8x = 24
x = 24/8
x = 3.....so the x int is 3 or (3,0)

now plot ur intercepts (3,0) and (0,12)......now start at ur y int (0,12)...and since ur slope is -4, u come down 4 spaces, then to the right 1 space, then down 4, and to the right 1...keep doing this and u should cross the x axis at (3,0)

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Answer:

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Step-by-step explanation:

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3 years ago

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Tanzania [10]

Try using Symbolab!

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3 years ago

Write an equation of the parabola that opens up whose vertex (−1, 2) is 3 units from the focus. (Vertex form or Parabola Form is

TEA [102]

Answer:

y2 = 4ax (opens right, a > 0)

y2 = -4ax (opens right, a > 0)

x2 = 4ay (opens up, a > 0)

x2 = -4ay (opens down, a > 0)

Vertex at (h, k) :

(y - k)2 = 4a(x - h) (opens right, a > 0)

(y - k)2 = -4a(x - h) (opens right, a > 0)

(x - h)2 = 4a(y - k) (opens up, a > 0)

(x - h)2 = -4a(y - k) (opens down, a > 0)

Equation of a Parabola in Vertex form

Vertex at Origin :

y = ax2 (opens up, a > 0)

y = -ax2 (opens down, a > 0)

x = ay2 (opens right, a > 0)

x = -ay2 (opens left, a > 0)

Vertex at (h, k) :

y = a(x - h)2 + k (opens up, a > 0)

y = -a(x - h)2 + k (opens down, a > 0)

x = a(y - k)2 + h (opens right, a > 0)

y = -a(y - k)2 + h (opens left, a > 0)

Step-by-step explanation:

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2 years ago

The ribbon was 3 1/4 inches wide and had an area of 19 1/2 square inches. What is the length of the ribbon? Please Help Will Giv

miv72 [106K]

Hello!

The formula for finding area is l × w

If the area is 19 ½ inches and it was 3 ¼ inches wide, then the length will be 19 ½ ÷ 3 ¼ (Area ÷ Width)

$19 \frac{1}{2} \div 3 \frac{1}{4}$

Change both to improper fraction

$19 \frac{1}{2} = \frac{39}{2}$

$3 \frac{1}{4} = \frac{13}{4}$

$\frac{39}{2} \div \frac{13}{4}$

Keep 39/2
Change ÷ to ×
Flip 13/4 = 4/13

$\frac{39}{2} \times \frac{4}{13} = \frac{156}{26} = 6$

$\framebox {The length of the ribbon is 6 inches}$

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2 years ago

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I really need help, this is due really soon. Just answer the question correctly and I will give brainy!

yanalaym [24]

Answer:

The graphs are the same because both equations have the same slope, the y-intercept only changes the place of the line. The lines are different because one line is at a different location than the other because of the difference in y intercepts.

Step-by-step explanation:

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3 years ago

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