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3 years ago

A line is parallel to y= 1/2x +6 and passes through (–8, 0). What is the equation of the line?

Mathematics

2 answers:

Harlamova29_29 [7]3 years ago

7 0

Answer:

y = 1/2x + 4

Step-by-step explanation:

y = 1/2x + b

0 = -4 + b

b = 4

y = 1/2x + 4

Bingel [31]3 years ago

7 0

keeping in mind that parallel lines have exactly the same slope, hmmm so what's the slope of the equation above?

$\bf y = \stackrel{\stackrel{m}{\downarrow }}{\cfrac{1}{2}}x+6\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

so, we're really looking for the equation of a line whose slope is 1/2 and runs through (-8 , 0)

$\bf (\stackrel{x_1}{-8}~,~\stackrel{y_1}{0})~\hspace{10em} \stackrel{slope}{m}\implies \cfrac{1}{2} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{\cfrac{1}{2}}[x-\stackrel{x_1}{(-8)}] \\\\\\ y=\cfrac{1}{2}(x+8)\implies y=\cfrac{1}{2}x+4$

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Using your equation of f(x)= 5x + 8, evaluate f(5)

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3 years ago

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Solve the system of equations. 10x+y=−20 y=2x2−4x−16 ( , ) and ( , )

Sladkaya [172]

I will go about solving this using the elimination method.

First, convert the equations.

10x + y = -20
4x + y = -12

Second, find the easiest variable to get rid of and get rid of it! (In this case, y) We will subtract to get rid of y.

6x = -8

Third, you want to solve the equation.

6x = -8 (divide by 6)
x = $-1 \frac{1}{3}$

Fourth, solve for y by inserting the answer for x into one of the equations.

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The solution for this system of equations is ( $-1 \frac{1}{3}$ , $-6 \frac{2}{3}$ ).

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3 years ago

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The population, P(t), of China, in billions, can be approximated by1 P(t)=1.394(1.006)t, where t is the number of years since th

vitfil [10]

Answer:

At the start of 2014, the population was growing at 8.34 million people per year.

At the start of 2015, the population was growing at 8.39 million people per year.

Step-by-step explanation:

To find how fast was the population growing at the start of 2014 and at the start of 2015 we need to take the derivative of the function with respect to t.

The derivative shows by how much the function (the population, in this case) is changing when the variable you're deriving with respect to (time) increases one unit (one year).

We know that the population, P(t), of China, in billions, can be approximated by $P(t)=1.394(1.006)^t$

To find the derivative you need to:

$\frac{d}{dt}\left(1.394\cdot \:1.006^t\right)=\\\\\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'\\\\1.394\frac{d}{dt}\left(1.006^t\right)\\\\\mathrm{Apply\:the\:derivative\:exponent\:rule}:\quad \frac{d}{dx}\left(a^x\right)=a^x\ln \left(a\right)\\\\1.394\cdot \:1.006^t\ln \left(1.006\right)\\\\\frac{d}{dt}\left(1.394\cdot \:1.006^t\right)=(1.394\cdot \ln \left(1.006\right))\cdot 1.006^t$

To find the population growing at the start of 2014 we say t = 0

$P(t)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^t\\P(0)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^0\\P(0)' = 0.00833901 \:Billion/year$

To find the population growing at the start of 2015 we say t = 1

$P(t)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^t\\P(1)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^1\\P(1)' = 0.00838904 \:Billion/year$