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krok68 [10]
3 years ago
13

In January, you deposit $150 into your checking account. Every month after that, you deposit 10% more than you did the month bef

ore. How much did you deposit in August?
Mathematics
2 answers:
just olya [345]3 years ago
7 0
Use a calculator and do 150*8 but I could be wrong so go with a different answer if there is one.
aliina [53]3 years ago
5 0
     The equation would be Y=15x+150 because 10% of 150 is 15. Next you coun up how many months you have up untill August and replace the x with it. So the equation would be Y=15(7)+150. ___Y=105+150___Y=$255. So $255 would be your answer.
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5 x + 4y= 16<br> X+y = 3.5<br><br> Find out what each letter is equal to <br> X=<br> Y=
klemol [59]
X=4
Y=-0.5

Do you need an explanation?
5 0
4 years ago
76<br> 34<br> х<br> Find the missing angle measures:<br> m ZX =<br> m ZY<br> m ZZ =
Veseljchak [2.6K]
Angle Y = 104 degrees
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7 0
4 years ago
Let f be a function defined for t ≥ 0. Then the integral ℒ{f(t)} = [infinity] e−stf(t) dt 0 is said to be the Laplace transform
sveticcg [70]

Answer:

L(f(t)) = 2 \frac{e^{-s} }{s} - \frac{1}{s}

Step-by-step explanation:

let f be a function defined for t ≥ 0

we can write the function f(t) in terms of unit function as follows

f(t) = 2 u,(t) - 1  where

      0≤ t < 1

f(t) = (2 * 0) -1 = -1

when t ≥ 1

f(t) = (2*1 )- 1 = 1

Now the Laplace transform L(F(T)) = 2L( u, (t) ) - L(1)  --------equation 1

this is because L(u,(t)) = \frac{e^{-cs} }{s}

c = 1 hence L(1) = 1/s

back to equation 1

L(f(t)) = 2 \frac{e^{-cs} }{s} -  1/s  laplace transform

also L(u(t) ) = \frac{e^{-s} }{s}

3 0
3 years ago
Can anyone explain how to get to the answer ? I’ll mark brainliest! Please
Svetllana [295]

Answer:

A

Step-by-step explanation:

It is linear because it is a pattern of adding 1 and not quadratic for there is no squaring. It is no exponential because the ratio stays the same.

7 0
4 years ago
PLEASE HELP! See attachment:
beks73 [17]
\bf f(x)=x^2e^{-2x}\\\\&#10;-----------------------------\\\\&#10;\cfrac{dy}{dx}=2xe^{-2x}+x^2\cdot -2e^{-2x}\implies \cfrac{dy}{dx}=xe^{-2x}(2-2x)&#10;\\\\\\&#10;\cfrac{dy}{dx}=\cfrac{x(2-2x)}{e^{2x}}\implies 0=\cfrac{x(2-2x)}{e^{2x}}&#10;\\\\\\&#10;0=x(2-2x)\implies &#10;\begin{cases}&#10;0=x\\&#10;0=2-2x\implies 2x=2\implies x=1&#10;\end{cases}

now, bear in mind, that zeroing out the denominator, also gives critical points, usually asymptotic points, where the derivative is undefined, now, in this case, the denominator is never zero, so we don't get any from the denominator, just from the numerator, and are 0 and 1

now check the picture below

running a first-derivative test on it, those are the values on those regions

you get a negative, regardless of what it might be, what matters is the sign
you get a positive, and then a negative

so, f(x) goes down, then up then down

now, you can see, there's on relative minimum and a relative maximum

6 0
3 years ago
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