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3 years ago

Katie got a loan for $10,000. The interest rate is 5%. How much interest will she pay in one year?

Mathematics

1 answer:

antiseptic1488 [7]3 years ago

5 0

Answer: 10,000 x 0.05 = $500

0.05 is equal to 5%

$500

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WILL MARK YOU BRAINLIEST

Nonamiya [84]

Answer:

_____________the answer is (-7,-3)______________

4 0

3 years ago

what if the ice caps were melting even more rapidly and san francisco was losing about 2% of its land area per year as land and

Helga [31]

The exponential function that shows San Francisco area after x years is y = 49(0.98)ˣ

<h3>What is an exponential function?</h3>

An exponential function is in the form:

y = abˣ

Where a is the initial value of y and b is the multiplication factor.

Let y represent the land area of San Francisco after x years. Hence:

The city is only about 49 square miles in area, a = 49. It is losing 2% yearly, hence b = 100% - 2% = 0.98

The equation becomes y = 49(0.98)ˣ

Find out more on exponential function at: brainly.com/question/12940982

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2 years ago

Use the variation of parameters method to solve the DR y" + y' - 2y = 1

postnew [5]

Answer:

$y(t)\ =\ C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

Step-by-step explanation:

As given in question, we have to find the solution of differential equation

$y"+y'-2y=1$

by using the variation in parameter method.

From the above equation, the characteristics equation can be given by

$D^2+D-2\ =\ 0$

$=>D=\ \dfrac{-1+\sqrt{1^2+4\times 2\times 1}}{2\times 1}\ or\ \dfrac{-1-\sqrt{1^2+4\times 2\times 1}}{2\times 1}$

$=>\ D=\ -2\ or\ 1$

Since, the roots of characteristics equation are real and distinct, so the complementary function of the differential equation can be by

$y_c(t)\ =\ C_1e^{-2t}+C_2e^t$

Let's assume that

$y_1(t)=e^{-2t}$ $y_2(t)=e^t$

$=>\ y'_1(t)=-2e^{-2t}$ $y'_2(t)=e^t$

and g(t)=1

Now, the Wronskian can be given by

$W=y_1(t).y'_2(t)-y'_1(t).y_2(t)$

$=e^{-2t}.e^t-e^t(-e^{-2t})$

$=e^{-t}+2e^{-t}$

$=3e^{-t}$

Now, the particular solution can be given by

$y_p(t)\ =\ -y_1(t)\int{\dfrac{y_2(t).g(t)}{W}dt}+y_2(t)\int{\dfrac{y_1(t).g(t)}{W}dt}$

$=\ -e^{-2t}\int{\dfrac{e^t.1}{3.e^{-t}}dt}+e^{t}\int{\dfrac{e^{-2t}.1}{3.e^{-t}}dt}$

$=\ -e^{-2t}\int{\dfrac{1}{3}dt}+\dfrac{e^t}{3}\int{e^{-t}dt}$

$=\dfrac{-e^{-2t}}{3}.t-\dfrac{1}{3}$

$=-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

Now, the complete solution of the given differential equation can be given by

$y(t)\ =\ y_c(t)+y_p(t)$

$=C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

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3 years ago

Simplify each expression -10-(-8)+(-4)-20

nata0808 [166]

-10- ( -8 ) + ( -4 ) - 20

= -10 + 8 - 4 - 20

= -2 - 4 - 20

= -6 - 20

= -26 <=== answer

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3 years ago

What is the Inverse equation of<br> (2x+1)/(3-2x)

STatiana [176]

The inverse means the opposite. So it would change all the signs.
(-2x-1)/(-3+2x)

6 0

4 years ago