A line the contains the point (-3,6) and (5,10)

Mathematics

1 answer:

wlad13 [49]3 years ago

5 0

Ok so for slope it's difference in y dividing by the difference of x. The difference of y is 10 - 6 which is 4, and the change in x is 5 - (-3) which is 8. Dividing 4 by 8 gets us 1/2. So the slope is 1/2. Slope intercept form would be y = mx + b, and the slope is 1/2 so it'd be y = (1/2)x + b. Plug in a point and you get b = 15/2. So the final slope intercept form would be y = (1/2)x + 15/2. For point slope form just put in a point and slope and you get y - 6 = (1/2) × (x + 3). Standard form would have to be arranged from slope intercept. So multiplying 2 to the slope intercept form we get 2y = x + 15. Subtracting x we get -x + 2y = 15. In standard form we don't want x to be negative so just multiply by negative one and we get x - 2y = -15.

You might be interested in

Solve the inequality and graph its solution: 5n - 10 > 25

Yanka [14]

Answer:

D. n>-7

Step-by-step explanation:

6 0

2 years ago

High expression is equivalent to ^4 square root x^10

Talja [164]

In order to find which expression is equivalent to $\sqrt[4]{x^{10}},$ simplify all given expressions:

$\sqrt[4]{x^{10}}=x^{\frac{10}{4}}=x^{\frac{5}{2}}=x^{2.5}.$

$x^2(\sqrt[4]{x^2})=x^2\cdot x^{\frac{2}{4}}=x^2\cdot x^{\frac{1}{2}}=x^2\cdot x^{0.5}=x^{2.5}.$

$x^{2.2}.$

$x^3(\sqrt[4]{x} )=x^3\cdot x^{\frac{1}{4}}=x^3\cdot x^{0.25}=x^{3.25}.$

$x^5.$

Therefore,

Answer: correct choice is A

7 0

3 years ago

Whats the intrgral of <img src="https://tex.z-dn.net/?f=%20%5Cint%20%20%5Cfrac%7Bx%5E2%2Bx-3%7D%7B%28x%5E3%2Bx%5E2-4x-4%29%5E2%7

rosijanka [135]

$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$

Notice that $x^3+x^2-4x-4=x^2(x+1)-4(x+1)=(x-2)(x+2)(x+1)$ . Decompose the integrand into partial fractions:

$\dfrac{x^2+x-3}{(x-2)^2(x+2)^2(x+1)^2}$
$=\dfrac1{3(x+1)}-\dfrac{11}{32(x+2)}-\dfrac1{3(x+1)^2}-\dfrac1{16(x+2)^2}+\dfrac1{96(x-2)}+\dfrac1{48(x-2)^2}$

Integrating term-by-term, you get

$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$
$=-\dfrac1{48(x-2)}+\dfrac1{3(x+1)}+\dfrac1{16(x+2)}+\dfrac1{96}\ln|x-2|+\dfrac13\ln|x+1|-\dfrac{11}{32}\ln|x+2|+C$