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2 years ago

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What is the equation of the line that has a slope of -4 and goes through (-1,6)? O A. y+ 6 = -4(x + 1) OB. y+6= -4(x - 1) OC. y-

6 = -4(x + 1) OD.y-6 = -4(x - 1)

1 answer:

Dennis_Churaev [7]2 years ago

4 0

Answer:

hi I dont now is this will help you but use M A T H W A Y it will give you the step b step info and the answere

Hope it help

Step-by-step explanation:

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What is the Y intercept pls help

Ad libitum [116K]

Answer:

y- intercept = 1

Step-by-step explanation:

The y- intercept is the point on the y- axis where the graph crosses.

This occurs when the x- coordinate is 0

From the table the point (0, 1 ) is where the graph crosses the y- axis

Then y- intercept = 1

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3 years ago

A mother explains to her child that the price on the sign is not the total price for an oil painting, because the price does not

dlinn [17]

Answer:

Or u cld simply just say 945 - 875 = 70

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2 years ago

50 POINTSS!!!!- The figure below shows the ideal pattern of movement of a herd of cattle, with the arrows showing the movement o

olga_2 [115]

Answer:

the first option

Step-by-step explanation:

solution:

distance: d=1/4 lc (because is a quarter of a circle)

length of the circumference: lc=2 pi r

pi: constant=3.14

radius of the circle: r=80 feet

replacing pi and r in the formula of lc:

lc=2 pi r

lc=2 (3.14) (80 feet)

lc=502.4 feet

replacing lc in the formula of d:

d=1/4 lc

d=1/4 (502.4 feet)

d=502.4/4 feet

d=125.6 feet

5 0

2 years ago

Read 2 more answers

HELP PLEASE EQUATION OF PARABOLA

Masteriza [31]

It has a minimum value at x = 3 and f(x) = 4

Vertex form is

f(x) = a(x - 3)^2 + 4 where a is some constant to be found

From the graph when x = 5 f(x) = 15, so

15 = a * 2^2 + 4

a = 15-4/4 = 11/4

so our equation is f(x) = 11/4(x - 3)^2 + 4

6 0

3 years ago

B. Determine the values of a and b so that f is continuous.<br><br> Please help!

wlad13 [49]

Answer:

following are the solution to this question:

Step-by-step explanation:

Please find the complete question in the attached file.

$\lim_{x\to 2}+f(x) = \lim_{x\to 2}+ (ax^2-bx+3) = 4a-2b+3\\\\ \to 4a-2b+3 = 4\\\\ \therefore \ \ 4a-2b = 1\\\\$

The one-sided limits of F(x) at x = 3 must be equivalent for f(x) to be continuous at x = 3.

$\to \lim_{x\to 3}- f(x) = \lim_{x\to 3}- (ax^2-bx+3) = 9a-3b+3\\\\ \to \lim_{x\to 3}+ f(x) = \lim_{x\to 3}+ (2x-a+b) = 6-a+b\\\\$

So,

$\to 9a-3b+3 = 6-a+b\\\\\therefore\\ \to 10a -4b = 3$

$\to 4a - 2b = 1......(a)\\\to 10a - 4b = 3.......(b)\\$

In equation a multiply the by -2 and then add in the equation b:

$\to -8a + 4b = -2\\\to 10a - 4b = 3\\ \to 2a = 1\\\\ \to a = \frac{1}{2}\\\\ \to \ 4(\frac{1}{2}) - 2b = 1\\\\ \to 2 - 2b = 1\\\\ \to -2b = -1\\\\ \to b = \frac{1}{2}$

So, the value of $a \ and \ b= \frac{1}{2}$

4 0

2 years ago