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3 years ago

7

Lou deposited $3,500 into a savings account. The simple interest rate is 4%. How much interest will the account earn in 10 years

?

1 answer:

finlep [7]3 years ago

3 0

Answer:

2 years : $280

10 years: $1400

Step-by-step explanation:

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Uncle Henry weights 183.1 pounds and aunt Wanda weights 139.24 pounds how much more does uncle Henry weight than aunt Wanda

sergij07 [2.7K]

Answer:

a

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

I took a picture of my math questions, here take a look.

PtichkaEL [24]

Which one do you want to be awnserd

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3 years ago

Alice works at a health care facility. She has measured the hemoglobin levels of 200 people. The data follows a normal distribut

amid [387]

This is a normal distribution with a Mean: 14 g/dL and a Standard deviation: 1 g/dL.
A ) Hemoglobin levels less than 13:
13 = 14 - 1 = Mean - 1 SD
0.16 x 200 = 32
Answer: 32 people.
B ) Hemoglobin levels greater than 14 :
0.50 x 200 = 100
Answer: 100 people.

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3 years ago

Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator

SpyIntel [72]

$\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}$
$\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)$

When $x=3$ , you're left with

$147=49a_1\implies a_1=\dfrac{147}{49}=3$

When $x=\sqrt2$ or $x=-\sqrt2$ , you're left with

$\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}$

Adding the two equations together gives $-10=2a_5$ , or $a_5=-5$ . Subtracting them gives $2\sqrt2=2\sqrt2a_4$ , $a_4=1$ .

Now, you have

$5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)$
$5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)$
$2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)$

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that $a_2x^4=2x^4$ and $a_3(-3)(-2)=6a_3=-6$ . These alone tell you that you must have $a_2=2$ and $a_3=-1$ .

So the partial fraction decomposition is

$\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}$

7 0

3 years ago

Divide <br> (6-q+8q^3-4q^2)/(2q-2)

mestny [16]

Answer:

Step-by-step explanation:

4q2 + 2q + 3

(2q - 2) l _ 8q3 - 4q2 - q + 6

8q3 - 8q2

_ 4q2 - q

4q2 - 4q

_ 3q + 6

6q + 6

-3q (remainder)

4q2 + 2q + 3 -3q / (2q - 2)

hope this helps

4 0

3 years ago