If you put $150 in a savings account that paid 6% compounded monthly, how much interest would you earn in five years?

Answers

Answer 1
Answer:

Answer:

Step-by-step explana8tion:

80$ with that 6%

Answer 2
Answer: Let me figure it out

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Given: Q = 7m + 3n, R = 11 - 2m, S = n + 5, and T = -m - 3n + 8.

Simplify Q + S - T.

Answers

Q+S-T=7m+3n+n+5-(-m-3n+8)\nQ+S-T=7m+4n+5+m+3n-8\nQ+S-T=8m+7n-3

Find the value
is the the correct answer

Answers

10^4 = (10)(10)(10)(10) = 10,000

What is the equation of the line that passes through the points (-2, 3) and (2, 7)?x - y = -1
x - y = -2
x - y = - 5
x - y = - 6

Answers

x -y = -5

To see that substitute the points.

First, point (-2,3)

-2 - 3 = -5

Now second point (2,7)

2 - 7 = -5

As you have seen, both points belong to the equation x - y = -5

Which of the following sums would be under the radical symbol to find the distance between the points (7, -1) and (-8, -9)?

Answers


Sadly, you didn't put any sums on your list of choices. 
In fact, I can't even find a list of choices !

To find the distance between  (7, -1)  and  (-8, -9),
you would build a sum under a square-root (radical)
symbol, and it would look something like this:

               √ ( 225 + 64 ) .
 

Evaluate\rm (3^2+1)/(3^2-1)+(5^2+1)/(5^2-1)+(7^2+1)/(7^2-1)+\ldots+(101^2+1)/(101^2-1) =
With step by step explanation !​

Answers

It's easier to deal with the symbolic sum (in sigma notation),

\displaystyle\sum_(k=1)^(50)((2k+1)^2+1)/((2k+1)^2-1)

Expanding the terms in the fraction, computing the quotient, and decomposing into partial fractions gives

((2k+1)^2+1)/((2k+1)^2-1) = (4k^2 + 4k + 2)/(4k^2 + 4k)

=\frac12*(2k^2 + 2k + 1)/(k^2 + k)

=\frac12\left(2+\frac1{k(k+1)}\right)

=\frac12\left(2 + \frac1k - \frac1{k+1}\right)

and it's the latter two terms that reveal a telescoping pattern.

In case you need more details about the partial fraction decomposition, we are looking for coefficients a and b such that

\frac1{k(k+1)}=\frac ak+\frac b{k+1}

or

1 = a(k+1) +bk =(a+b)k+a

which gives a = 1, and a + b = 0 so that b = -1.

Our sum has been rearranged as

\displaystyle\frac12\sum_(k=1)^(50)\left(2+\frac1k-\frac1{k+1}\right)=\sum_(k=1)^(50)1+\frac12\sum_(k=1)^(50)\left(\frac1k-\frac1{k+1}\right)=50+\frac12\sum_(k=1)^(50)\left(\frac1k-\frac1{k+1}\right)

The remaining telescoping sum is

1/2 [(1/1 - 1/2) + (1/2- 1/3) + (1/3- 1/4) + … + (1/48- 1/49) + (1/49- 1/50) + (1/50 - 1/51)]

and you can see how there are pairs of numbers that cancel, so that the sum reduces to

1/2 [1/1 - 1/51] = 1/2 [1 - 1/51] = 1/2 × 50/51 = 25/51

So, our original sum ends up being

\displaystyle\sum_(k=1)^(50)((2k+1)^2+1)/((2k+1)^2-1) = 50 + (25)/(51) = \boxed{(2575)/(51)}

In the drawing below, PQ is parallel to ST. Triangle PQR is similar to triangle TSR. Which statement about the sides is true?

Answers

Answer:

PR corresponds to TR

Step-by-step explanation:

<Q and <S are pair of alternate angles [since PQ is parallel to ST]

<Q = <S [since they are alternate angles]

So, PR corresponds to TR [since they are opposite sides of alternate angles and ΔPQR is similar to ΔTSR]

Answer:

PR corresponds to TR.

Step-by-step explanation:

Its the second choice:

PR corresponds to TR.

This is because they are opposite equal angles ( < Q and <S).