2 years ago

Given the points (0,−3) and (5,-6) and on a line, find its equation in the form y=mx+b

Mathematics

1 answer:

Andrew [12]2 years ago

4 0

Answer:

Step-by-step explanation:

y=mx+b

b=-3 (y intercept)

Use equation y2-y1/x2-x1

-3/5= m

y=-3/5x-3

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The sum of 237 and six times the opposite of a number is 405. What is the number?

Bond [772]

Answer:

(A) x=-28

Step-by-step explanation:

237 -6x = 405 [note that 6*(-x) is (-6x)]

-6x = 168 [subtract 237 from both sides]

x = -28 [divide both sides by (-6)]

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2 years ago

Given below are some inequalities. Plot the feasible region graphically.

igor_vitrenko [27]

Answer:

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Step-by-step explanation:

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Write in numeric form. Nine hundred thirty six and twelve hundredths

4vir4ik [10]

That would be... 936.12

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Write a definite integral that represents the area of the region. (Do not evaluate the integral.) y1 = x2 + 2x + 3 y2 = 2x + 12F

Svet_ta [14]

Answer:

$A = \int\limits^3__-3}{9}-{x^{2}} \, dx = 36$

Step-by-step explanation:

The equations are:

$y = x^{2} + 2x + 3$

$y = 2x + 12$

The two graphs intersect when:

$x^{2} + 2x + 3 = 2x + 12$

$x^{2} = 0$

$x_{1} = 3\\x_{2} = -3$

To find the area under the curve for the first equation:

$A_{1} = \int\limits^3__-3}{x^{2} + 2x + 3} \, dx$

To find the area under the curve for the second equation:

$A_{2} = \int\limits^3__-3}{2x + 12} \, dx$

To find the total area:

$A = A_{2} -A_{1} = \int\limits^3__-3}{2x + 12} \, dx -\int\limits^3__-3}{x^{2} + 2x + 3} \, dx$

Simplifying the equation:

$A = \int\limits^3__-3}{2x + 12}-({x^{2} + 2x + 3}) \, dx = \int\limits^3__-3}{9}-{x^{2}} \, dx$

Note: The reason the area is equal to the area two minus area one is that the line, area 2, is above the region of interest (see image).

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9x – 7 = 9x + h<br> What dose h equal to

deff fn [24]

Answer:

h= -7

hope this helps!!:)

Step-by-step explanation:

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