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2 years ago

(2x+1)(x^2+8x-3) help standard form

Mathematics

1 answer:

trapecia [35]2 years ago

6 0

Answer:

$2x^3+17x^2+2x-3$

Step-by-step explanation:

$\left(2x+1\right)\left(x^2+8x-3\right)\\=2xx^2+2x\cdot \:8x+2x\left(-3\right)+1\cdot \:x^2+1\cdot \:8x+1\cdot \left(-3\right)\\=2x^3+17x^2+2x-3$

In this equation, the variables were isolated to simplify the problem, following the order of operations. Negative-three either became an exponent or positive integer, 3, subtracted by a byproduct of the previous equation.

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Answer:

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3 years ago

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Taya2010 [7]

The bag contains,

Red (R) marbles is 9, Green (G) marbles is 7 and Blue (B) marbles is 4,

Total marbles (possible outcome) is,

$\text{Total marbles = (R) + (G) +(B) = 9 + 7 + 4 = 20 marbles}$

Let P(R) represent the probablity of picking a red marble,

P(G) represent the probability of picking a green marble and,

P(B) represent the probability of picking a blue marble.

Probability , P, is,

$\text{Prob, P =}\frac{required\text{ outcome}}{possible\text{ outcome}}$ $\begin{gathered} P(R)=\frac{9}{20} \\ P(G)=\frac{7}{20} \\ P(B)=\frac{4}{20} \end{gathered}$

Probablity of drawing a Red marble (R) and then a blue marble (B) without being replaced,

That means once a marble is drawn, the total marbles (possible outcome) reduces as well,

$\begin{gathered} \text{Prob of a red marble P(R) =}\frac{9}{20} \\ \text{Prob of }a\text{ blue marble =}\frac{4}{19} \\ \text{After a marble is selected without replacement, marbles left is 19} \\ \text{Prob of red marble + prob of blue marble = P(R) + P(B) = }\frac{9}{20}+\frac{4}{19}=\frac{251}{380} \\ \text{Hence, the probability is }\frac{251}{380} \end{gathered}$

Hence, the best option is G.

5 0

1 year ago

(2x+1)(x^2+8x-3) help standard form​

(2x+1)(x^2+8x-3) help standard form