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OLga [1]
3 years ago
12

Which of the following lines is parallel to x = 8?

Mathematics
1 answer:
Ulleksa [173]3 years ago
5 0
Interestingly enough, the line x=8 is parallel to the y axis.  The line x=8 means that x=8 no matter what y equals.  
So any line that is parallel to the y axis could be the answer.

x=4
x=12
x=-17
x=-2.5   are all examples of lines that are parallel to x=8.  Perhaps that will answer your question.
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Which of these statements are true?
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Step-by-step explanation:

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Pls solve for x!: 2x−5.6 over 3 = 1−x over 1.5
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Step-by-step explanation:

1 2/3 (5 2/7)

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A college has a 30% completion rate, meaning that 30% of all students who start at the college complete the goal they set. The p
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45%

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For simplicity, let use assume there are 100 students in the school.

No. of students to complete college = (30/100) x 100 = 30 Students

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Therefore;

Rate goal % = (45/100) x 100% = 45%

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Layla went shopping for camping equipment. After looking at several different stores, she bought a tent for $163.63 and a sleepi
Ghella [55]

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$297.09

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To answer this question, just add:

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So, Layla spent $297.09 in all

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5 0
3 years ago
Kryptonite is a material found on the planet Krypton and has various effects, most importantly on Superman. The most common type
Rashid [163]

Answer:

The maximum amount of red kryptonite present is 33.27 g after 4.93 hours.

Step-by-step explanation:

dy/dt = y(1/t - k)

separating the variables, we have

dy/y = (1/t - k)dt

dy/y = dt/t - kdt

integrating both sides, we have

∫dy/y = ∫dt/t - ∫kdt

㏑y = ㏑t - kt + C

㏑y - ㏑t = -kt + C

㏑(y/t) = -kt + C

taking exponents of both sides, we have

\frac{y}{t} = e^{-kt + C}  \\\frac{y}{t} = e^{-kt}e^{C} \\\frac{y}{t} = Ae^{-kt}   (A = e^{C})\\y = Ate^{-kt}

when t = 1 hour, y = 15 grams. So,

y = Ate^{-kt}\\15 = A(1)e^{-kX1}\\15 = Ae^{-k}(1)

when t = 3 hours, y = 30 grams. So,

y = Ate^{-kt}\\30 = A(3)e^{-kX3}\\30 = 3Ae^{-3k} (2)

dividing (2) by (1), we have

\frac{30}{15}  = \frac{3Ae^{-3k}}{Ae^{-k}} \\2 = 3e^{-2k}\\\frac{2}{3} = e^{-2k}

taking natural logarithm of both sides, we have

-2k = ㏑(2/3)

-2k = -0.4055

k = -0.4055/-2

k = 0.203

From (1)

A = 15e^{k} \\A = 15e^{0.203} \\A = 15 X 1.225\\A = 18.36

Substituting A and k into y, we have

y = 18.36te^{-0.203t}

The maximum value of y is obtained when dy/dt = 0

dy/dt = y(1/t - k) = 0

y(1/t - k) = 0

Since y ≠ 0, (1/t - k) = 0.

So, 1/t = k

t = 1/k

So, the maximum value of y is obtained when t = 1/k = 1/0.203 = 4.93 hours

y = 18.36(1/0.203)e^{-0.203t}\\y = \frac{18.36}{0.203}e^{-0.203X1/0.203}\\y = 90.44e^{-1}\\y = 90.44 X 0.3679\\y = 33.27 g

<u>So the maximum amount of red kryptonite present is 33.27 g after 4.93 hours.</u>

3 0
3 years ago
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