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3 years ago

How to solve systems of equations

1 answer:

7 0

You are given two equations, solve for one variable in one of the equations. Say you solved for x in the second equation. Then, plug in that value of x in the x of the first equation. Solve this (first) equation for y (as it should become apparent) and you'll get a number value. Plug in this numerical value of y into the y of the second equation. Solve for x in the second equation. And there you have it: (x, y)

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The average score of all pro golfers for a particular course has a mean of 70 and a standard deviation

Lisa [10]

Answer:

Step-by-step explanation:

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2 years ago

7/8 x c for c =8 what is the value

ikadub [295]

If the x stands for multiplying then it will equal 7. take 7 divided by 8 and then multiply that by 8 and you get seven. did that help my dear?

7 0

3 years ago

Read 2 more answers

1+2=21, 2+3=36,3+4=43,4+5=?

Anni [7]

1+2=21, 2+3=36, 3+4=43, 4+5=52.

4 0

3 years ago

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Determine the values of c so that the following functions represent joint probability distributions of the random variables X an

Anna11 [10]

Answer:

The joint probility density will be c = 1/36

The joint probability distribution will be c = 1/15

Step-by-step explanation:

For step by step explanation see the pucture attached

7 0

3 years ago

Read 2 more answers

Find the domain of the function y = 3 tan(23x)

solmaris [256]

Answer:

$\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

In other words, the $x$ in $f(x) = 3\, \tan(23\, x)$ could be any real number as long as $x \ne \displaystyle \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)$ for all integer $k$ (including negative integers.)

Step-by-step explanation:

The tangent function $y = \tan(x)$ has a real value for real inputs $x$ as long as the input $x \ne \displaystyle k\, \pi + \frac{\pi}{2}$ for all integer $k$ .

Hence, the domain of the original tangent function is $\mathbb{R} \backslash \displaystyle \left\lbrace \left. \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

On the other hand, in the function $f(x) = 3\, \tan(23\, x)$ , the input to the tangent function is replaced with $(23\, x)$ .

The transformed tangent function $\tan(23\, x)$ would have a real value as long as its input $(23\, x)$ ensures that $23\, x\ne \displaystyle k\, \pi + \frac{\pi}{2}$ for all integer $k$ .

In other words, $\tan(23\, x)$ would have a real value as long as $x\ne \displaystyle \frac{1}{23} \, \left(k\, \pi + \frac{\pi}{2}\right)$ .

Accordingly, the domain of $f(x) = 3\, \tan(23\, x)$ would be $\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

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2 years ago