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3 years ago

Use the graph of the polynomial function to find the factored form of the

Mathematics

1 answer:

Fofino [41]3 years ago

4 0

Answer:

Step-by-step explanation:

when factoring, the two numbers that you get after you factor are also called solutions, or x intercepts. from this, we know that since the line crosses at positive 1 and positive 9, the answer must be -1 and -9 as you flip the sign after factoring to get the real answer

(ps you can also just plug in all of the answer choices to desmos and see which one matches your graph)

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I need help with this one

timama [110]

Answer:

A, it is the 3rd line in a set of 8. 1/8 equals .125 Multiply 3X.125 and it equals .375

Step-by-step explanation:

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4 years ago

Please someone help me on how to do grids?

mafiozo [28]

Answer:

25

Step-by-step explanation:

Question: The distance between P and T on the coordinate grid is __ units. (Input whole numbers only.)

Answer + Step-by-step explanation:

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3 years ago

Read 2 more answers

78% of the class completed their homework last night. What fraction of the class completed their homework?

mixas84 [53]

Answer:

$\frac{39}{50}$

Step-by-step explanation:

=> 78% makes:

=> $\frac{78}{100}$

<u><em>In simplest form:</em></u>

=> $\frac{39}{50}$

3 0

3 years ago

Read 2 more answers

Write out the first few terms of the series Summation from n equals 0 to infinity (StartFraction 2 Over 3 Superscript n EndFract

anyanavicka [17]

Answer:

$\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n} = 15/8$

Step-by-step explanation:

The sum you are trying to understand is this.

$\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n}$

Remember that in general when you have a geometric series

$\sum\limits_{n = 0}^{\infty} a*r^n$ you have that

$\sum\limits_{n = 0}^{\infty} a*r^n = \frac{a}{1-r}$ and that equality is true as long as $|r| < 1$ .

Therefore here we have

$\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n} = \sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{3*5} \big)^n = \sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{15} \big)^n$ and $\big|\frac{-1}{15} \big| = \frac{1}{15} < 1$