What expression is equivalent to 2(9+7)?

The way to work this out is to find a common denominator. so in this case 80 is a common denominator. So it would be 3/8 into 30/80 and then 4/10 into 32/80 therefore 4/10 is bigger

7 0

2 years ago

Find an equation of a line passing through the point (8,9) and parallel to the line joining the points (2,7) and (1,5).

TiliK225 [7]

Answer:

2x - y - 7 = 0

Step-by-step explanation:

Since the slope of parallel line are same.

So, we can easily use formula,

y - y₁ = m ( x ₋ x₁)

where, (x₁, y₁) = (8, 9)

and m is a slope of line passing through (x₁, y₁).

and since the slope of parallel lines are same, so here we use slope of parallel line for calculation.

and, Slope = m = $\dfrac{y_{b}-y_{a}}{x_{b}-x_{a}}$

here, (xₐ, yₐ) = (2, 7)

and, $(y_{a},y_{b}) = (1, 5 )$

⇒ m = $\dfrac{5-7}{1-2}$

⇒ m = 2

Putting all values above formula. We get,

y - 9 = 2 ( x ₋ 8)

⇒ y - 9 = 2x - 16

⇒ 2x - y - 7 = 0

which is required equation.

3 0

2 years ago

Read 2 more answers

What is the antiderivative of 3x/((x-1)^2)

Maslowich

Answer:

$\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C$

Step-by-step explanation:

Given

$\int \:\:3\cdot \frac{x}{\left(x-1\right)^2}dx$

$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$=3\cdot \int \frac{x}{\left(x-1\right)^2}dx$

$\mathrm{Apply\:u-substitution:}\:u=x-1$

$=3\cdot \int \frac{u+1}{u^2}du$

$\mathrm{Expand}\:\frac{u+1}{u^2}:\quad \frac{1}{u}+\frac{1}{u^2}$

$=3\cdot \int \frac{1}{u}+\frac{1}{u^2}du$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$=3\left(\int \frac{1}{u}du+\int \frac{1}{u^2}du\right)$

as

$\int \frac{1}{u}du=\ln \left|u\right|$ ∵ $\mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)$

$\int \frac{1}{u^2}du=-\frac{1}{u}$ ∵ $\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1$

so

$=3\left(\ln \left|u\right|-\frac{1}{u}\right)$

$\mathrm{Substitute\:back}\:u=x-1$

$=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)$

$\mathrm{Add\:a\:constant\:to\:the\:solution}$

$=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C$

Therefore,

$\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C$

4 0

3 years ago