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3 years ago

Help me please I’ll give BRAINLYIST. A person runs 1 mile in

Mathematics

1 answer:

Crank3 years ago

4 0

Answer:

since speed is miles per hour it would be s=1/1/2 so it would be 2mph

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Use the distributive property to remove the parentheses? -8(-4w-3v+6)

olganol [36]

Answer:

32w + 24v - 48

Step-by-step explanation:

- 8(- 4w - 3v + 6) ← multiply each term in the parenthesis by - 8

= 32w + 24v - 48

4 0

2 years ago

To write the polynomial 5x^4+8x^5-2x+3 in standard form, which term is the leading term?

Nana76 [90]

The answer is B. The leading coefficient is always the variable with the highest degree. In this case, it's 5.

8 0

3 years ago

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Monica worked 3.5 hours and is paid $10 per hour. How much did she make in 3.5 hours?

marysya [2.9K]

Answer:

She would make $35.

Step-by-step explanation:

If you do 10 x 3.5, you will get the answer of $35.

7 0

3 years ago

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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

NO LINKS!! Please help me with these notes. Part 1a

erik [133]

Answers:

When we evaluate a logarithm, we are finding the exponent, or power x, that the base b, needs to be raised so that it equals the argument m. The power is also known as the exponent.

$5^2 = 25 \to \log_5(25) = 2$