Question 9 (1 point)

Mathematics

1 answer:

Luden [163]3 years ago

8 0

Answer:

Step-by-step explanation:

16x-80+18=4x+22

16x-62=4x+22

12x-62=22

12x=84

x=7

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one possible combination of homeroom groups is 18 groups of 10 students and 5 groups of 12 students. which equation uses the x =

vfiekz [6]

Answer: y - 5 = -5/6 (x - 18)

Step-by-step explanation:

The point-slope form of a linear equation is written using the slope of the line and one point in the line. From part A, the slope of the line representing this situation is m = -5/6.

Since x represents the number of 10-student groups and y represents the number of 12-student groups, the combination of 18 groups of 10 students and 5 groups of 12 students is represented by the point (18,5).

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3 years ago

Product of (y^3 - 125) ÷ (y - 5)

Ludmilka [50]

Answer:

$y^{2}+5y+25$

Step-by-step explanation:

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3 years ago

Each week, Alfonso earns $85 for every shift he works at the theater and $15 for every dog-walking job. He uses the expression 8

sesenic [268]

So you didn't put an actual question. To find the money he makes you just plug in the numbers.
If he works 4 shifts at the theater you replace x with 4. Then, you find 85 x 4. You do the same thing with 15y.

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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,

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