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andreyandreev [35.5K]
3 years ago
11

HELLPP PLEASE :) THIS IS HARD BRAINLIEST!

Mathematics
1 answer:
lorasvet [3.4K]3 years ago
7 0

Answer:

1).B

2).A

3).D

4).B

5).C

6).D

7).D

8).B

9).C

10). I believe it is A

Step-by-step explanation:

I'm sorry, i do not know how to explain this, I hope I helped you in some way and I hope these answers work for you :)

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3 years ago
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Find the distance from the point \left(-8,-3,10\right) to the plane 3 x - 8 y+0 z=-8.
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Diana put $8000 in a 10-year CDpaying 5% interest compoundedmonthly. After 2 years, shewithdrew all her money, and as anearly wi
Svet_ta [14]

Answer:

$8,430.23

Explanation:

From the statement of the problem:

• The principal amount = $8,000

,

• Interest Rate = 5%

,

• Compounding Period = 12 (Monthly)

The compound interest formula is given as:

A(n)=P\left(1+\frac{r}{k}\right)^{nk}

Using the compound period formula, we first, calculate the amount in her account at the end of 1 year.

\begin{gathered} A(1)=8000\left(1+\frac{0.05}{12}\right)^{12\times1} \\ A(1)=\$8409.30 \end{gathered}

This means that the interest she made during the first year is:

\text{ Interest during the first year}=8409.30-8000=\$409.30

Next, calculate the amount in her account at the end of the second year.

\begin{gathered} A(2)=8000\left(1+\frac{0.05}{12}\right)^{12\times2} \\ A(2)=\$8839.53 \end{gathered}

Since she paid back all the interest she made during the first year, the amount Diana was left with is:

\begin{gathered} 8839.53-409.30 \\ =8,430.23 \end{gathered}

Diana was left with $8,430.23.

6 0
1 year ago
Describe the behavior of the function ppp around its vertical asymptote at x=-2x=−2x, equals, minus, 2. ​
insens350 [35]

Answer:

x->-2^{-}, p(x)->-\infty and as x->-2^{+}, p(x)->-\infty

Step-by-step explanation:

Given

p(x) = \frac{x^2-2x-3}{x+2} -- Missing from the question

Required

The behavior of the function around its vertical asymptote at x = -2

p(x) = \frac{x^2-2x-3}{x+2}

Expand the numerator

p(x) = \frac{x^2 + x -3x - 3}{x+2}

Factorize

p(x) = \frac{x(x + 1) -3(x + 1)}{x+2}

Factor out x + 1

p(x) = \frac{(x -3)(x + 1)}{x+2}

We test the function using values close to -2 (one value will be less than -2 while the other will be greater than -2)

We are only interested in the sign of the result

----------------------------------------------------------------------------------------------------------

As x approaches -2 implies that:

x -> -2^{-} Say x = -3

p(x) = \frac{(x -3)(x + 1)}{x+2}

p(-3) = \frac{(-3-3)(-3+1)}{-3+2} = \frac{-6 * -2}{-1} = \frac{+12}{-1} = -12

We have a negative value (-12); This will be called negative infinity

This implies that as x approaches -2, p(x) approaches negative infinity

x->-2^{-}, p(x)->-\infty

Take note of the superscript of 2 (this implies that, we approach 2 from a value less than 2)

As x leaves -2 implies that: x>-2

Say x = -2.1

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We have a negative value (-56.1); This will be called negative infinity

This implies that as x leaves -2, p(x) approaches negative infinity

x->-2^{+}, p(x)->-\infty

So, the behavior is:

x->-2^{-}, p(x)->-\infty and as x->-2^{+}, p(x)->-\infty

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3 years ago
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kirill115 [55]

Answer:

Step-by-step explanation:

Minimum: 2

Quartile Q1: 4.5

Median: 10

Quartile Q3: 15

Maximum: 19

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