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3 years ago

Complete the square to transform the quadratic equation into the form (x – p)2 = q.

1 answer:

3 0

If this problem is expecting you to show a graph photo, then a dependable website for a problem like this would be www.desmos.com. message me if you're not sure how to use it.

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HELP ASAP!!!<br> Find the value of 'A' in the set of complementary angles.

dem82 [27]

answer:

180 - 63 = 117

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3 years ago

You have averaged $62,500.00 for the last five years of work. your company has agreed to pay you a yearly pension of 75 %. what

fenix001 [56]

$600,400,145..........

7 0

3 years ago

HELP MEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

goldfiish [28.3K]

Setup is
y = k/x
8 = k/3
k = 24

Now use 6 for y
6 = 24/x
x = 24/6 = 4 or A

5 0

3 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

the brain volumes () of 50 brains vary from a low of to a high of . use the range rule of thumb to estimate the standard devia

Sloan [31]

The estimated standard deviation from the range rule is 145.5 cm3.As it is 29.8 cm3 smaller than the exact standard deviation, we consider that the estimation is not accurate enough.

Generally, brain volumes which are smaller than 921.1 cm3 can be considered significantly low and bigger than 1417.5 cm3 can be considered significantly high. Hence, brain volume of 1457.5 cm3 is bigger than the expected maximum, so it can be considered significantly high.

The calculation goes like this,

According to the range rule, the standard deviation and range, or the distance between the maximum and smallest value, are roughly inversely proportional:

$s=R/4$

If we have a sample of 50 brains volumes, and they range from 910 cm3 to 1492 cm3, we can estimate the standard deviation as:

$s=max-min/4$

s=1492-910/4=582/4=145.5

The exact standard deviation is 175.3 cm3.The difference is 175.3-145.5=29.8 cm3, so the estimation is not accurate enough.

The range rule of thumb states that we should anticipate a maximum value that is approximately two standard deviations above the mean and a lowest value that is approximately two standard deviations below the mean. The expected ranges are as follows given our mean volume of 1169.3 cm3 and standard deviation of 124.1 cm3:

Min= M-2s=1169.3-2*124.1=921.1

Max=M+2s=1169.3+2*124.1=1417.5

The brain capacity of 1457.5 cm3 can be regarded as significantly high because it is higher than the predicted maximum.

To know about standard deviation click here :

brainly.com/question/23907081

#SPJ4

5 0

1 year ago