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3 years ago

Line L passes through the points (1, 1) and (7, 8). What is the slope of a line that is perpendicular to line L?

1 answer:

8 0

When a line passes through the two points $(x_{1},y_{1}) and (x_{2},y_{2})$ , its slope is given by the formula $m= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

In this question, a line L passes through the points $(1,1) and (7,8)$

So, its slope is given by $m=\frac{8-1}{7-1}$

$m=\frac{7}{6}$

When two lines are perpendicular, then the product of their slopes is -1.

Since, the slope of the line L is $\frac{7}{6}$ , so the slope of the line which is perpendicular to the given line L is $\frac{-6}{7}$ as the product of $\frac{-6}{7} \times \frac{7}{6}=-1$ .

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Write the fractions in lowest terms:

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Part 1:

A and B are already in lowest terms

C 3/8

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Part 2:

A 5 * 11

B 3 * 3 * 3

C 2 * 2 * 3 * 5

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4 years ago

Find x in this 45°-45°-90° triangle.<br><br> x=

rosijanka [135]

Answer:

wow that looks hard hope someone helps u

6 0

3 years ago

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Use the bionomial theorem to write the binomial expansion

MaRussiya [10]

Answer:

$\left(\frac{x}{2} + 3 y\right)^{4}=\frac{x^{4}}{16} + \frac{3}{2} x^{3} y + \frac{27}{2} x^{2} y^{2} + 54 x y^{3} + 81 y^{4}$

Step-by-step explanation:

$\left(\frac{1}{2}x+3y \right)^4=\left(\frac{x}{2}+3y \right)^4\\$

Binomial Expansion Formula:

$(a+b)^n=\sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$ , also $\binom{n}{k}=\frac{n!}{(n-k)!k!}$

We have to solve $\left(\frac{x}{2} + 3 y\right)^{4}=\sum_{k=0}^{4} \binom{4}{k} \left(3 y\right)^{4-k} \left(\frac{x}{2}\right)^k$

Now we should calculate for $k=0, k=1, k=2, k=3 \text{ and } k =4;$

First, for $k=0$

$\binom{4}{0} \left(3 y\right)^{4-0} \left(\frac{x}{2}\right)^{0}=\frac{4!}{(4-0)! 0!}\left(3 y\right)^{4} \left(\frac{x}{2}\right)^{0}=\frac{4!}{4!}(81y^4)\cdot 1 =81 y^{4}$

It is the same procedure for the other:

For $k=1$

$\binom{4}{1} \left(3 y\right)^{4-1} \left(\frac{x}{2}\right)^{1}=54 x y^{3}$

For $k=2$

$\binom{4}{2} \left(3 y\right)^{4-2} \left(\frac{x}{2}\right)^{2}=\frac{27}{2} x^{2} y^{2}$

For $k=3$

$\binom{4}{3} \left(3 y\right)^{4-3} \left(\frac{x}{2}\right)^{3}=\frac{3}{2} x^{3} y$

For $k=4$

$\binom{4}{4} \left(3 y\right)^{4-4} \left(\frac{x}{2}\right)^{4}=\frac{x^{4}}{16}$

You can perform the calculations, I will not type everything.

The answer is the sum of elements calculated.

Just organizing:

$\left(\frac{x}{2} + 3 y\right)^{4}=\frac{x^{4}}{16} + \frac{3}{2} x^{3} y + \frac{27}{2} x^{2} y^{2} + 54 x y^{3} + 81 y^{4}$

8 0

3 years ago

Read 2 more answers

Help asap!!! compute $\left(-3\sqrt{128}\right)\left(-4\sqrt{50}\right)$

labwork [276]

$(-3\sqrt{128})(-4\sqrt{50})\qquad \begin{cases} 128=2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\\ \qquad 2^6\cdot 2\\ \qquad (2^3)^2\cdot 2\\ 50=5\cdot 5\cdot 2\\ \qquad 5^2\cdot 2 \end{cases} \\\\\\ (-3\sqrt{(2^3)^2\cdot 2})(-4\sqrt{5^2\cdot 2})\implies 12\sqrt{(2^3)^2\cdot 2}\cdot \sqrt{5^2\cdot 2} \\\\\\ 12\cdot 2^3\sqrt{2}\cdot 5\sqrt{2}\implies 12\cdot 2^3\cdot 5\cdot \sqrt{2^2}\implies 12\cdot 2^3\cdot 5\cdot 2\implies 960$

5 0

3 years ago

The population of Jose’s town in 1995 was 2400 and the population in 2000 was

Gnom [1K]

Answer:

$y=320x+2400$

Step-by-step explanation:

Remember, a line has the form $y=mx+b$ , where $m$ is the slope and $b$ is the y-intercept. In this cases, y represents the population when have been passed x years since 1995.

Since x represent the number of years since 1995, then in 1995, x=0 and the population is 2400. Then a point of the linear equation is (x,y)=(0,2400).

In 2000 has been passes 5 years since 1995 and the population is 4000, this means that x=5 and y=4000. Then other point in the line is (x,y)=(5,4000)

For find the equation of the line that describe the preview situation we use the points found.

Remember, if $(x_1, y_1)$ and $(x_2,y_2)$ are pointsin the plane, the slope of the line that passes through the points is