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alekssr [168]
2 years ago
10

Write an equation for each line : m = -4 and the y-intercept is 3

Mathematics
1 answer:
QveST [7]2 years ago
3 0

Answer:

y = -4x + 3

Step-by-step explanation:

m = slop

b = y-int

y = mx + b

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I've tried solving this but I'm stuck -8=2/x
USPshnik [31]

Answer:

-0.25

Step-by-step explanation:

-8= 2/x

multiply both sides by x -> -8x =2

divide both sides by -8 -> x=2/-8

Ans= -1/4 = -0.25

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3 years ago
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PLEASE PLEASE HELP ME 30 POINTS!<br><br> x + 20 ≤ -10
Artemon [7]

X + 20 <= -10

Subtract 20 from both sides:

X <= -30

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3 years ago
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Evaluate the following expressions: 2(−1 + 3) − 7
Svetach [21]

Answer:

-3 is the answer.

Step-by-step explanation:

=2(-1+3)-7

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3 years ago
A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of wat
jeyben [28]

Answer:

N(1)=50 is a minimum

N(15)=4391.7 is a maximum

Step-by-step explanation:

<u>Extrema values of functions </u>

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.  

We are given a function (corrected)

N(t) = 20(t^2-lnt^2)+ 30

N(t) = 20(t^2-2lnt)+ 30

(a)

First, we take its derivative

N'(t) = 20(2t-\frac{2}{t})

Solve N'(t)=0

20(2t-\frac{2}{t})=0

Simplifying

2t^2-2=0

Solving for t

t=1\ ,t=-1

Only t=1 belongs to the valid interval 1\leqslant t\leqslant 15

Taking the second derivative

N''(t) = 20(2+\frac{2}{t^2})

Which is always positive, so t=1 is a minimum

(b)

N(1)=20(1^2-2ln1)+ 30

N(1)=50 is a minimum

(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15

(d)

N(15)=20(15^2-2ln15)+ 30

N(15)=4391.7 is a maximum

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3 years ago
Which part of the decision-making process is directly related to brainstorming?
Salsk061 [2.6K]

C.Generating possible option's.

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