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2 years ago

7

What is the slope of a line that is perpendicular to the graph of y = 2x + 5? A-² B 72 C2 D-2

1 answer:

Greeley [361]2 years ago

7 0

The correct answer it’s C

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(a-b)(a^2 + ab + b^2)

Nitella [24]

Answer:

a^3-b^3

Step-by-step explanation:

(a−b)(a2+ab+b2)

=(a+−b)(a2+ab+b2)

=(a)(a2)+(a)(ab)+(a)(b2)+(−b)(a2)+(−b)(ab)+(−b)(b2)

=a3+a2b+ab2−a2b−ab2−b3

=a3−b3

6 0

2 years ago

Read 2 more answers

Calculate the rate of change.

liq [111]

Answer:

It is 5/5 or 1/1 as it goes up 5 and to the left 5!!!

5 0

2 years ago

7. Triangle FIP was dilated by a scale factor of centered at the origin to

Assoli18 [71]

Answer:

AB is parallel to A'B'.

DO,1/2 (1/2x, 1/2y) =

The distance from A' to the origin is half the distance from A to the origin.

Step-by-step explanation:

6 0

2 years ago

A line has a slope of -2/5 and passes through the point (10,7) . Write the equation of this line in point-slope form, slope-inte

qaws [65]

$(\stackrel{x_1}{10}~,~\stackrel{y_1}{7})\hspace{10em} \stackrel{slope}{m} ~=~ -\cfrac{2}{5} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{7}=\stackrel{m}{-\cfrac{2}{5}}(x-\stackrel{x_1}{10})$

6 0

1 year ago

Two streams flow into a reservoir. Let X and Y be two continuous random variables representing the flow of each stream with join

zlopas [31]

Answer:

c = 0.165

Step-by-step explanation:

Given:

f(x, y) = cx y(1 + y) for 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3,

f(x, y) = 0 otherwise.

Required:

The value of c

To find the value of c, we make use of the property of a joint probability distribution function which states that

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, dy \, dx = 1$

where a and b represent -infinity to +infinity (in other words, the bound of the distribution)

By substituting cx y(1 + y) for f(x, y) and replacing a and b with their respective values, we have

$\int\limits^3_0 \int\limits^3_0 {cxy(1+y)} \, dy \, dx = 1$

Since c is a constant, we can bring it out of the integral sign; to give us

$c\int\limits^3_0 \int\limits^3_0 {xy(1+y)} \, dy \, dx = 1$

Open the bracket

$c\int\limits^3_0 \int\limits^3_0 {xy+xy^{2} } \, dy \, dx = 1$

Integrate with respect to y

$c\int\limits^3_0 {\frac{xy^{2}}{2} +\frac{xy^{3}}{3} } \, dx (0,3}) = 1$

Substitute 0 and 3 for y

$c\int\limits^3_0 {(\frac{x* 3^{2}}{2} +\frac{x * 3^{3}}{3} ) - (\frac{x* 0^{2}}{2} +\frac{x * 0^{3}}{3})} \, dx = 1$

$c\int\limits^3_0 {(\frac{x* 9}{2} +\frac{x * 27}{3} ) - (0 +0) \, dx = 1$

$c\int\limits^3_0 {(\frac{9x}{2} +\frac{27x}{3} ) \, dx = 1$

Add fraction

$c\int\limits^3_0 {(\frac{27x + 54x}{6}) \, dx = 1$

$c\int\limits^3_0 {\frac{81x}{6} \, dx = 1$

Rewrite;

$c\int\limits^3_0 (81x * \frac{1}{6}) \, dx = 1$

The $\frac{1}{6}$ is a constant, so it can be removed from the integral sign to give

$c * \frac{1}{6}\int\limits^3_0 (81x ) \, dx = 1$

$\frac{c}{6}\int\limits^3_0 (81x ) \, dx = 1$

Integrate with respect to x

$\frac{c}{6} * \frac{81x^{2}}{2} (0,3) = 1$

Substitute 0 and 3 for x

$\frac{c}{6} * \frac{81 * 3^{2} - 81 * 0^{2}}{2} = 1$

$\frac{c}{6} * \frac{81 * 9 - 0}{2} = 1$

$\frac{c}{6} * \frac{729}{2} = 1$

$\frac{729c}{12} = 1$

Multiply both sides by $\frac{12}{729}$

$c = \frac{12}{729}$

$c = 0.0165 (Approximately)$

8 0

3 years ago