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Alex777 [14]
4 years ago
10

What is the point used in the equation of the line y+4=1/2(x-2)

Mathematics
1 answer:
Ket [755]4 years ago
8 0

Answer:

What is the point used in the equation of the line y+4=1/2(x-2)

The other format for straight-line equations is called the "point-slope" form. For this one, they give you a point (x1, y1) and a slope m, and have you plug it into this formula:

y - y1 = m(x - x1)

Don't let the subscripts scare you. They are just intended to indicate the point they give you. You have the generic "x" and generic "y" that are always in your equation, and then you have the specific x and y from the point they gave you; the specific x and y are what is subscripted in the formula. Here's how you use the point-slope formula

They've given me m = 4, x1 = -1, and y1 = -6.  I'll plug these values into the point-slope form, and solve for "y=":

y - y1 = m(x - x1)

y - (-6) = (4)(x - (-1))

y + 6 = 4(x + 1)

y + 6 = 4x + 4

y = 4x + 4 - 6

y = 4x - 2 

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How are the expressions 8-15 and 8-(-15)?how are they different?
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The concentration of a drug in the bloodstream C(t) at any time t, in hours, is described by the equationC(t)=100t/t^2+25where t
alisha [4.7K]

Answer:

It will take 5 hours until it reaches its maximum concentration.

Step-by-step explanation:

The maximum concentration will happen in t hours. t is found when

C'(t) = 0

In this problem

C(t) = \frac{100t}{t^{2} + 25}

Applying the quotient derivative formula

C'(t) = \frac{(100t)'(t^{2} + 25) - (t^{2} + 25)'(100t)}{(t^{2} + 25)^{2}}

C'(t) = \frac{100t^{2} + 2500 - 200t^{2}}{(t^{2} + 25)^{2}}

C'(t) = \frac{-100t^{2} + 2500}{(t^{2} + 25)^{2}}

A fraction is equal to zero when the numerator is 0. So

-100t^{2} + 2500 = 0

100t^{2} = 2500

t^{2} = 25

t = \pm \sqrt{25}

t = \pm 5

We use only positive value.

It will take 5 hours until it reaches its maximum concentration.

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3 years ago
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Answer:

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Step-by-step explanation:

Please see the attached picture for the full solution.

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