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3 years ago

Which is the slope-intercept form of an equation for the line containing

Mathematics

1 answer:

Thepotemich [5.8K]3 years ago

6 0

Answer:

The answer is letter A.

Step-by-step explanation:

y=mx+b

-3=(-1)(0)+b

-3=0+b

0 0

-3=b

y= -1x-3

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Which is the least common multiple of 4 and 12

spayn [35]

The answer is 12. My explanations is that LCM is the smallest number that both numbers can fit into. 12 fits into 12 once, and 4 fits into 12 3 times, so 12 is the answer.

3 0

2 years ago

Help me plzzzzzzzzzzzzzzzzzz.

poizon [28]

Answer:

<u>Option A </u>is your answer

Step-by-step explanation:

hope it helps you

have a great day

6 0

3 years ago

Can I get help I'm very confused

horrorfan [7]

Answer:

y=(-3/4)x-3

The question is asking you under what condition does x must go through to equal to y according to this graph; like what formula is the used on the input (x) to equal the output (y)

3 0

2 years ago

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modelled by the function C(t)=8(e

Alexxx [7]

Answer:

the maximum concentration of the antibiotic during the first 12 hours is 1.185 $\mu g/mL$ at t= 2 hours.

Step-by-step explanation:

We are given the following information:

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modeled by the function where the time t is measured in hours and C is measured in $\mu g/mL$

$C(t) = 8(e^{(-0.4t)}-e^{(-0.6t)})$

Thus, we are given the time interval [0,12] for t.

We can apply the first derivative test, to know the absolute maximum value because we have a closed interval for t.
The first derivative test focusing on a particular point. If the function switches or changes from increasing to decreasing at the point, then the function will achieve a highest value at that point.

First, we differentiate C(t) with respect to t, to get,

$\frac{d(C(t))}{dt} = 8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)})$

Equating the first derivative to zero, we get,

$\frac{d(C(t))}{dt} = 0\\\\8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0$

Solving, we get,

$8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0\\\displaystyle\frac{e^{-0.4}}{e^{-0.6}} = \frac{0.6}{0.4}\\\\e^{0.2t} = 1.5\\\\t = \frac{ln(1.5)}{0.2}\\\\t \approx 2$

At t = 0

$C(0) = 8(e^{(0)}-e^{(0)}) = 0$

At t = 2

$C(2) = 8(e^{(-0.8)}-e^{(-1.2)}) = 1.185$

At t = 12

$C(12) = 8(e^{(-4.8)}-e^{(-7.2)}) = 0.059$

Thus, the maximum concentration of the antibiotic during the first 12 hours is 1.185 $\mu g/mL$ at t= 2 hours.

4 0

3 years ago

Help again lol

olga2289 [7]

For the number before x you would look for the rise over the run. in this case you can see that there is an intersection at (-2,0) and (0,1) you find the difference ox each so you have 2/1 or just 2 you can tell that the line is going up so it is positive. for the addition part look at where the y intercept is in this case it is at positive one. this makes your equation y=2x+1

8 0

3 years ago

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