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3 years ago

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Which of the following could be the equation of the function below? On a coordinate plane, a curve crosses the y-axis at y = 2.

It has a maximum at 5 and a minimum at negative 1. It goes through one cycle at pi. y = negative 3 sine (2 (x + pi)) + 2 y = negative 3 sine (x + pi) + 2 y = 3 sine (4 (x minus pi)) + 4 y = 3 sine (2 (x + pi)) + 2

2 answers:

Mashcka [7]3 years ago

7 0

Answer:

b

Step-by-step explanation:

Neko [114]3 years ago

3 0

Answer:

I believe according to the information provided that the answer would be would be B

Step-by-step explanation:

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(2n-9)-(-2.4n+4)Find the difference.

raketka [301]

Answer:

The difference is 4.4n−13.

Step-by-step explanation:

7 0

3 years ago

Which relations are functions?<br><br><br><br> Choose all answers that are correct.

-Dominant- [34]

Hello!

In a function, each input has only one output. In A, three has two outputs, 4 and 5, so A is not a function.

In B, you can use something called the vertical line test to see if each x value has one y value as an output. You move an imaginary vertical line across the graph, and if it intersects with two points it is not a function. If we do this on our graph, it will not intersect two points. Therefore, B is a function.

In C, we can see that each input has one output, or there are all different inputs, so C is a function.

For D we can use that vertical line test again. It intersects both the points (-1,1) and (-1,6) so D is not a function

Our final answers are B and C.

I hope this helps!

6 0

3 years ago

Read 2 more answers

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

Jayden just graduated from college and owes $19,400 on his student loans. The

ELEN [110]

Answer:

M=19400(1.5)/1-(1+1.5)-(15x12)

M=28917.50

Then you would round up so it wo9uld be $28,918.

6 0

3 years ago

What is 76.58 plus 24.3 equallys to what? Please explain how you added and got your answer

Varvara68 [4.7K]

I just added the two numbers and got the number 100.88

6 0

3 years ago

Read 2 more answers