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Aleksandr [31]
3 years ago
11

You invest $1,300 in an account with an annual interest rate of 2.5%, compounded monthly. How much money is in the account after

15 years? Round your answer to the nearest whole number.
Mathematics
1 answer:
wariber [46]3 years ago
4 0
Hi there
The formula is
A=p (1+r/k)^kt
A future value?
P present value 1300
R interest rate 0.025
K compounded monthly 12
T time 15 years
So
A=1,300×(1+0.025÷12)^(12×15)
A=1,890.75

Good luck!
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What can you say about the continuous function that generated the following table of values?
scoundrel [369]

Option C. The function that has the table here has  exactly one x-intercept.

<h3>How to find the intercept</h3>

As the signs of the functions are changing, we can see that there is a crossing to the x axis.

This tells us that the zero value that we see is the x intercept and there is no other x intercept in the question.

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brainly.com/question/2500907

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5 0
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Which of the following choices is the standard deviation of the sample shown here? 19,20,21,22,23
serg [7]

Answer:

Answer : 42

Step-by-step explanation:

<h2>Hope this help :)</h2>

6 0
3 years ago
Can I have your help​
Volgvan

Answer:

3/5

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sin is opposite divided by the hypotenuse ( longest side)

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4 years ago
Read 2 more answers
Let u = (1, k) and v = (2, 1). Find k such that The distance between u and v is 1 u and v are orthogonal The angle between u and
SVETLANKA909090 [29]

Answer:k=1,k=-2,k=8\pm 5\sqrt{3}

Step-by-step explanation:

Given two vectors

u=1\hat{i}+k\hat{j}

v=2\hat{i}+1\hat{j}

\left ( i\right )Distance between them is given by

|u-v|=\sqrt{\left ( 2-1\right )^2+\left ( 1-k\right )^2}=1

squaring both side

1^{2}+\left ( 1-k\right )^2=1

k^2-2k+1=0

\left ( k-1\right )^2=0

k=1

\left ( ii\right )

angle between u and v is 90 i.e. orthogonal

u\dot v=0

\left ( 1\hat{i}+k\hat{j}\right )\dot \left ( 2\hat{i}+1\hat{j}\right )=0

2+k=0

k=-2

\left ( iii\right )

angle between u & v is \frac{\pi }{3}

u\dot v=|u||v|cos\left (\frac{\pi }{3}\right )

|u|=\sqrt{1^2+k^2}

|v|=\sqrt{2^2+1^2}

2+k=\left ( \sqrt{1+k^2}\right )\left ( \sqrt{5}\right )cos\left ( \frac{\pi }{3}\right )

\left ( 4+2\right )^2=\left ( 1+k^2\right )5

k^2-16k-11=0

k=8\pm 5\sqrt{3}

7 0
3 years ago
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