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2 years ago

Use long division to divide x3 + x2 - 2x + 14 by x + 3. Choose the correct quotient.

Mathematics

1 answer:

True [87]2 years ago

3 0

Answer:

x2 - 2x + 4 (remainder 2).

or we could write it as

x2 - 2x + 4 + 2/(x+3)

Step-by-step explanation:

x2 - 2x + 4 <------ quotient.

x + 3 ) x3 + x2 - 2x + 14

x3 + 3x2

-2x^2 - 2x

-2x2 - 6x

4x + 14

4x + 12

2 <--------- remainder

Step-by-step explanation:

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Tanzania [10]

C is the correct answer

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3 years ago

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Which phrase best describes the transformation from the graph of y=(2x-15)^2+3 to the graph of y=2(x-11)^2+3?

weeeeeb [17]

Answer:

The graph's vertex would move 4 to the right and would become twice as wide.

Step-by-step explanation:

We can tell this because the x value of the vertex is the constant inside the parenthesis. In this case, the number changes from 11 to 15, which means it moves to the right by 4.

And with the 2 outside of the parenthesis instead of inside, it becomes more wide (since the 2 is not being squared).

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2 years ago

When a bactericide is added to a nutrient broth in which bacteria are growing, the bacteria population continues to grow for a

baherus [9]

Answer:

a) 1296 bacteria per hour

b) 0 bacteria per hour

c) -1296 bacteria per hour

Step-by-step explanation:

We are given the following information in the question:

The size of the population at time t is given by:

$b(t) = 9^6 + 6^4t-6^3t^2$

We differentiate the given function.

Thus, the growth rate is given by:

$\displaystyle\frac{db(t)}{dt} = \frac{d}{dt}(9^6 + 6^4t-6^3t^2)\\\\= 6^4-2(6^3)t$

a) Growth rates at t = 0 hours

$\displaystyle\frac{db(t)}{dt} \bigg|_{t=0}= 6^4-2(6^3)(0) = 1296\text{ bacteria per hour}$

b) Growth rates at t = 3 hours

$\displaystyle\frac{db(t)}{dt} \bigg|_{t=3}= 6^4-2(6^3)(3) = 0\text{ bacteria per hour}$

c) Growth rates at t = 6 hours

$\displaystyle\frac{db(t)}{dt} \bigg|_{t=6}= 6^4-2(6^3)(6) = -1296\text{ bacteria per hour}$

3 0

3 years ago

The expression cosine of pi over 2 times cosine of pi over 3 plus sine of pi over 2 times sine of pi over 3 can be rewritten as

9966 [12]

Solution:

Given;

$\cos\frac{\pi}{2}\cos\frac{\pi}{3}+\sin\frac{\pi}{2}\sin\frac{\pi}{3}$

Using the identities;

Thus;

$\begin{gathered} \cos\frac{\pi}{2}\cos\frac{\pi}{3}+\sin\frac{\pi}{2}\sin\frac{\pi}{3}=\cos(\frac{\pi}{2}-\frac{\pi}{3}) \\ \\ \cos\frac{\pi}{2}\cos\frac{\pi}{3}+\sin\frac{\pi}{2}\sin\frac{\pi}{3}=\cos(\frac{\pi}{6}) \end{gathered}$

CORRECT OPTION: A

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1 year ago

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Gemiola [76]

Answer:

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3 years ago