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3 years ago

Find and x and y PLEASE HELP

Mathematics

1 answer:

Bogdan [553]3 years ago

6 0

Answer:

Step-by-step explanation:

x is 79 and y is 47

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A mix of factoring by grouping. Factor the exressions. 17ax+17a-x-1 Please explain/show steps on how to do this!

mario62 [17]

Answer:

$(x+1)(17a-1)$

Step-by-step explanation:

We are given that

$17ax+17a-x-1$

We have to factorize the given expression.

Pair up the terms

$(17ax+17a)+(-x-1)$

Now, taking common factor

$17a(x+1)-1(x+1)$

Taking common (x+1)

$(x+1)(17a-1)$

Now, we get

$(x+1)(17a-1)$

This is required factors of given expression.

4 0

3 years ago

Segment A'B' is parallel to segment AB. What is the length of segment AB?

Nonamiya [84]

Answer:

7.5

Step-by-step explanation:

5/6=x/9

cross multiply

6x=45

x=45/6

x=7 3/6 = 7 1/2 = 7.5

7 0

3 years ago

The number of miles Ford trucks can go on one tank of gas is normally distributed with a mean of 350 miles and a standard deviat

kati45 [8]

Answer:

The probability that a randomly chosen Ford truck runs out of gas before it has gone 325 miles is 0.0062.

Step-by-step explanation:

Let X = the number of miles Ford trucks can go on one tank of gas.

The random variable X is normally distributed with mean, μ = 350 miles and standard deviation, σ = 10 miles.

If the Ford truck runs out of gas before it has gone 325 miles it implies that the truck has traveled less than 325 miles.

Compute the value of P (X < 325) as follows:

$P(X$

Thus, the probability that a randomly chosen Ford truck runs out of gas before it has gone 325 miles is 0.0062.

7 0

3 years ago

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

Mazyrski [523]

We can use the fact that, for $|x|$ ,

$\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n$

Notice that

$\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1{1-x}\right]=\dfrac1{(1-x)^2}$

so that

$f(x)=\displaystyle\frac5{(1-x)^2}=5\frac{\mathrm d}{\mathrm dx}\left[\sum_{n=0}^\infty x^n\right]$

$f(x)=\displaystyle5\sum_{n=0}^\infty nx^{n-1}$

$f(x)=\displaystyle5\sum_{n=1}^\infty nx^{n-1}$

$f(x)=\displaystyle5\sum_{n=0}^\infty(n+1)x^n$

By the ratio test, this series converges if

$\displaystyle\lim_{n\to\infty}\left|\frac{(n+2)x^{n+1}}{(n+1)x^n}\right|=|x|\lim_{n\to\infty}\frac{n+2}{n+1}=|x|$

so the series has radius of convergence $R=1$ .

5 0

3 years ago

Which of these pairs of events are dependent?

Pani-rosa [81]

Answer:

the first one

Step-by-step explanation:

8 0

3 years ago