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2 years ago

Write an equation of the line passing through the point A(2, 0) that is parallel to the line y = 3x−5 .

1 answer:

4 0

Hello,

In order for lines to be parallel to each other, they need to have the same slope. So, let's plug in the points and slope for point-slope form.

The equation for point-slope form is: y-y1 = m (x-x1)

Let's plug in the points into the equation.
y-5 = m (x-0), and then let's plug in the slope for "m" : y-5 = 3 (x-0).

After plugging the numbers in, the equation is simplified to y-5 = 3 (x-0). Then distribute the 3 to the parenthesis. y-5 = 3x. To get y by itself, you need to add 5 to both sides. After that, the equation parallel to that equation is: y = 3x + 5.

Hope this helps! =)

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3. Is the relationship shown by the data linear? If so, model the data with an equation X | Y 1 ,5| 5,10| 9,15| 13,20|

son4ous [18]

Answer:

$y=\frac{5}{4}x+\frac{15}{4}$

Step-by-step explanation:

x | y

-------

1 5

5 10

9 15

13 20

This one is linear because as x goes up by the same number so does y. So the ratio of difference of y to difference of x is the same per pair of points.

So a linear equation in slope-intercept form is y=mx+b where m is the slope and b is the y-intercept.

To find the slope, I'm going to line up the points vertically and subtract, then put 2nd difference over 1st difference. Like so,

( 1 , 5)

-( 5 , 10)

----------------

-4 -5

So the slope is 5/4 which makes sense since the y's are going up by 5 each time and the x's are going up by 4 each time.

So we have m=5/4. Let's plug that into our y=mx+b.

y=5/4 x+b

To find b, we need to use y=5/4 x+b along with one of the given points.

Choose; it doesn't matter. I like (1,5) I guess.

y=5/4 x +b with (1,5)

5=5/4 (1)+b

5=5/4 +b

Subtract 5/4 on both sides:

5-5/4=b

20/4-5/4=b (Found a common denominator)

15/4=b

The y-intercept is 15/4 so b=15/4.

So the equation for the line in slope-intercept form is y=5/4 x +15/4.

$y=\frac{5}{4}x+\frac{15}{4}$

4 0

3 years ago

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Alex777 [14]

The answer is 84 3/8 because 672 and the left over 3 so it would be 84 3/8.

Hope I Helped!!!!!!!!

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2 years ago

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Amiraneli [1.4K]

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Step-by-step explanation: i did it before

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2 years ago

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asambeis [7]

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7 0

2 years ago

Enter the coefficients of the fifth Taylor polynomial T5(x) for the function f(x) = x5−3x4+2x2+5x−2 based at b=1. T5(x)= + (x−1)

DENIUS [597]

Compute the necessary values/derivatives of $f(x)$ at $x=1$ :

$f(1)=3$

$f'(1)=2$

$f''(1)=-12$

$f'''(1)=-12$

$f^{(4)}(1)=48$

$f^{(5)}(1)=120$

Taylor's theorem then says we can "approximate" (in quotes because the Taylor polynomial for a polynomial is another, exact polynomial) $f(x)$ at $x=1$ by

$T_5(x)=\dfrac3{0!}+\dfrac2{1!}(x-1)-\dfrac{12}{2!}(x-1)^2-\dfrac{12}{3!}(x-1)^3+\dfrac{48}{4!}(x-1)^4+\dfrac{120}{5!}(x-1)^5$

$T_5(x)=3+2(x-1)-6(x-1)^2-2(x-1)^3+2(x-1)^4+(x-1)^5$

###

Another way of doing this would be to solve for the coefficients $a,b,c,d,e,g$ in

$f(x)=a+b(x-1)+c(x-1)^2+d(x-1)^3+e(x-1)^4+g(x-1)^5$

by expanding the right hand side and matching up terms with the same power of $x$ .

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2 years ago