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4 years ago

10

Complete the point-slope equation of the line through (2,3 and (7,4). Use exact numbers.

1 answer:

uysha [10]4 years ago

3 0

The point-slope equation of the line is $y-3=\frac{1}{5}(x-2)$

Step-by-step explanation:

The form of the point-slope equation is $y-y_{1}=m(x-x_{1})$ , where

m is the lope of the line
$(x_{1},y_{1})$ is a point lies on the line

The slope of a line $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ , where

$(x_{1},y_{1})$ and $(x_{2},y_{2})$ are two points on the line

∵ The line through (2 , 3) and (7 , 4)

∴ $x_{1}$ = 2 and $x_{2}$ = 7

∴ $y_{1}$ = 3 and $y_{2}$ = 4

- Substitute these value in the rule of the slope

∵ $m=\frac{4-3}{7-2}=\frac{1}{5}$

∴ the slope of the line is $m=\frac{1}{5}$

Let us substitute the value of the slope and the coordinates of point $(x_{1},y_{1})$ in the form of the equation

∵ $y-y_{1}=m(x-x_{1})$

∵ $x_{1}$ = 2 and $y_{1}$ = 3

∵ $m=\frac{1}{5}$

∴ $y-3=\frac{1}{5}(x-2)$

The point-slope equation of the line is $y-3=\frac{1}{5}(x-2)$

Learn more:

You can learn more about the linear equation in brainly.com/question/12941985

#LearnwithBrainly

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Flauer [41]

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3 years ago

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Dafna1 [17]

Answer:7

Step-by-step explanation:

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3 years ago

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4 years ago

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Makovka662 [10]

Let $u=x^2-4$ and $v=4x-5$ . By the product rule,

$\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=\dfrac{\mathrm d(u^5)}{\mathrm dx}v^4+u^5\dfrac{\mathrm d(v^4)}{\mathrm dx}$

By the power rule, we have $(u^5)'=5u^4$ and $(v^4)'=4v^3$ , but $u,v$ are functions of $x$ , so we also need to apply the chain rule:

$\dfrac{\mathrm d(u^5)}{\mathrm dx}=5u^4\dfrac{\mathrm du}{\mathrm dx}$

$\dfrac{\mathrm d(v^4)}{\mathrm dx}=4v^3\dfrac{\mathrm dv}{\mathrm dx}$

and we have

$\dfrac{\mathrm du}{\mathrm dx}=2x$

$\dfrac{\mathrm dv}{\mathrm dx}=4$

So we end up with

$\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=10xu^4v^4+16u^5v^3$

Replace $u,v$ to get everything in terms of $x$ :

$\dfrac{\mathrm d((x^2-4)^5(4x-5)^4)}{\mathrm dx}=10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3$

We can simplify this by factoring:

$10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3=2(x^2-4)^4(4x-5)^3\bigg(5x(4x-5)+8(x^2-4)\bigg)$

$=2(x^2-4)^4(4x-5)^3(28x^2-57)$

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3 years ago

Draw the image of the following segment after a dilation centered at the origin with a scale factor of 2/3

Irina18 [472]

Answer:

Step-by-step explanation:

Let the ends of the given segment are A and B.

Coordinates of A → (8, 6)

Coordinates of B → (12, 12)

If a point (x, y) is dilated by a scale factor 'k' about the origin, rule to be followed,

(x, y) → (kx, ky)

If k = $\frac{2}{3}$

(x, y) → $(\frac{2}{3}x, \frac{2}{3}y)$

By this rule coordinates of the image points of A and B will be,

A(8, 6) → $A'(\frac{2\times 8}{3},\frac{2\times 6}{3})$

→ A'(5.3, 4)

B(12, 12) → $B'(\frac{12\times 2}{3}, \frac{12\times 2}{3})$

→ B'(8, 8)

Now we can get the image of segment AB after dilation by a scale factor of $\frac{2}{3}$ .

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3 years ago