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3 years ago

Which is the complete factorization of this expression -25x-25

Mathematics

1 answer:

Annette [7]3 years ago

7 0

Factor −25x−25

−25x−25

=−25(x+1)

Answer:

−25(x+1)

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Help!! Can someone tell me how to do this with work shown?!

kykrilka [37]

Add all the numbers up
thats how u get perimeter
so, 28+21+11+46+19
add them and u gettt....
125 and add the unit which is feet and all together it is
125 ft.

7 0

3 years ago

Somebody said they are all false. is that true?

mestny [16]

Answer: B

Step-by-step explanation:

The question asks for [n≥-5] to be true, which means any value that is greater or equal to -5 will be considered.

A. {-5, -5.5, -6}

-5 TRUE

-5.5 FALSE

-6 FALSE

B. {-5, -4.5, -3}

-5 TRUE

-4.5 TRUE

-3 TRUE

C. {-6, 0, 5}

-6 FALSE

0 TRUE

5 TRUE

D. {-6, -7, -8}

-6 FALSE

-7 FALSE

-8 FALSE

As shown above, we can clearly see that Set B is the set that makes the inequality true.

8 0

3 years ago

Courtney hit the Bullseye 3/5 of the time when playing darts Write 3/5 as a Decimal

Tema [17]

Answer: 0.6

Step-by-step explanation: you divide 3 by 5

3 0

3 years ago

Identify the equation modeled below,and then find its solution

Usimov [2.4K]

Answer:

Step-by-step explanation:

4x-4=-x-1 = 3/5 or .6

8 0

3 years ago

Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i

jenyasd209 [6]

Answer:

If the limit that you want to find is $\lim_{x\to \infty}\dfrac{P(x)}{Q(x)}$ then you can use the following proof.

Step-by-step explanation:

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ and $Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0}$ be the given polinomials. Then

$\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}$

Observe that

$\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}$

and

$\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}$

Then

$\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}$

3 0

3 years ago