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3 years ago

Write the equation of the line in slope-intercept form.

Mathematics

1 answer:

finlep [7]3 years ago

7 0

Answer
y=mx+b
y=2x-7

Explanation
m = the “common difference” between two points
In this graph, it’s 2 because every time the line moves across the x axis by 1, it increases in 2 in the y axis

b = the point that intercepts the y axis; the y-intercept
In this graph, the point intercepts the y axis at -7 (the y-intercept is -7)

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Which of the following functions has the same horizontal asymptote as the function graphed below?

Alla [95]

Answer:

In most cases, there are two types of functions that have horizontal asymptotes. Functions in quotient form whose denominators are bigger than numerators when x is large positive or large negative.

Step-by-step explanation:

7 0

3 years ago

WILL GIVE 100 POINTS AND BRAINLIEST FOR BEST ANSWER

coldgirl [10]

Answer:

Step-by-step explanation:-16x^2 + 24x + 16 = 0.

A. Divide by 8:

-2x^2 + 3x + 2 = 0, A*C = -2*2 = -4 = -1 * 4. Sum = -1 + 4 = 3 = B, -2x^2 + (-x+4x) + 2 = 0,

(-2x^2-x) + (4x+2) = 0,

-x(2x+1) + 2(2x+1) = 0,

(2x+1)(-x+2) = 0, 2x+1 = 0, X = -1/2. -x+2 = 0, X = 2.

X-intercepts: (-1/2,0), (2,0).

B. Since the coefficient of x^2 is negative, the parabola opens downward. Therefore, the vertex is a maximum.

Locate the vertex: h = Xv = -B/2A = -24/-32 = 3/4, Plug 3/4 into the given Eq to find k(Yv). K = -16(3/4)^2 + 16(3/4) + 16 = 19. V(h,k) = V(3/4,19).

C. Choose 3 points above and below the vertex for graphing. Include the points calculated in part A which shows where the graph crosses the x-axis.

4 0

3 years ago

In MON, J, K, and L are midpoints. If JL = 11, LK = 13, and ON = 20, and JL || MN, LK || MO, and JK || ON, what is the length of

In-s [12.5K]

Answer:

The lengths of MN is 22 units, MO is 26 units and JK is 10 units

Step-by-step explanation:

A line segment joining the mid-points of two sides in a triangle is parallel to the third side and equal to half its length

In Δ MON

∵ J, K, and L are mid-points

∵ JL // MN and LK // MO

∴ L is the mid-point of ON

∴ J is the mid-point of MO

∴ K is the mid-point of MN

∵ J, L are the mid-points of MO and ON

∵ JL is opposite to MN

→ By using the rule above

∴ JL = $\frac{1}{2}$ MN

∵ JL = 11 units

∴ 11 = $\frac{1}{2}$ MN

→ Multiply both sides by 2

∴ 22 = MN

∴ MN = 22 units

∵ K, L are the mid-points of MN and ON

∵ KL is opposite to MO

→ By using the rule above

∴ KL = $\frac{1}{2}$ MO

∵ KL = 13 units

∴ 13 = $\frac{1}{2}$ MO

→ Multiply both sides by 2

∴ 26 = MO

∴ MO = 26 units

∵ J, K are the mid-points of MO and MN

∵ JK is opposite to ON

→ By using the rule above

∴ JK = $\frac{1}{2}$ ON

∵ ON =20 units

∴ JK = $\frac{1}{2}$ (20)

∴ JK = 10 units

∴ The lengths of MN are 22 units, MO is 26 units and JK is 10 units

3 0

3 years ago

Addition fact of 12-4

Fiesta28 [93]

I Think It Is 8. Onstaly

8 0

3 years ago

Read 2 more answers

Which of the following inequalities is graphed on the coordinate plane

azamat

Answer:

The answer is D.

Step-by-step explanation:

Well, all the equations are the same except for the symbol so you don't need to figure out the equation of the line.

Anyways, note that the line is dotted, which means it cannot be A nor B since they have an "equal to."

Also, note that the shaded region is above the graph. Since it is above, it means that y is greater than whatever x is.

Thus, the answer is D.

5 0

3 years ago