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2 years ago

5

Victor has $90,000 in a savings account that earns 14% annually. The interest is not compounded.How much will he have in 4 month

s?

1 answer:

TiliK225 [7]2 years ago

6 0

Answer:

$ 5200

Step-by-step explanation:

int = money . time . rate

int = 90000 . 4/12 . 14/100

int = 900 . 1/3 . 14

int = 300 . 14

int = 5200

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Work out the percentage change when a price of £80 is increased to £100.

Mars2501 [29]

Answer:

125%

Step-by-step explanation:

80/100=8

80*x=100

100/80=1.25

1.25*100=125

x=125%

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2 years ago

Find the slope of the line that passes through (9,8) and (4,1)

katen-ka-za [31]

Answer:

1.4

I hope this helps!

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2 years ago

What is 5 & 3/4% as a decimal

8_murik_8 [283]

5 and 3/5 is 5.75 as a decimal

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2 years ago

Question 2 - Signal and System Properties. State whether each of the statements is true or false. Note that a statement is true

gregori [183]

Answer:

(a) true

(b) true

(c) false; {y = x, t < 1; y = 2x, t ≥ 1}

(d) false; y = 200x for .005 < |x| < 1

Step-by-step explanation:

(a) "s(t) is periodic with period T" means s(t) = s(t+nT) for any integer n. Since values of n may be of the form n = 2m for any integer m, then this also means ...

s(t) = s(t +2mt) = s(t +m(2T)) . . . for any integer m

This equation matches the form of a function periodic with period 2T.

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(b) A system being linear means the output for the sum of two inputs is the sum of the outputs from the separate inputs:

s(a) +s(b) = s(a+b) . . . . definition of linear function

Then if a=b, you have

2s(a) = s(2a)

__

(c) The output from a time-shifted input will only be the time-shifted output of the unshifted input if the system is time-invariant. The problem conditions here don't require that. A system can be "linear continuous time" and still be time-varying.

__

(d) A restriction on an input magnitude does not mean the same restriction applies to the output magnitude. The system may have gain, for example.

6 0

3 years ago

Find all the values of x in the set of complex numbers that satisfy the following equation:

Fynjy0 [20]

Compute all the component integrals first:

$I_1=\displaystyle\int_0^{\pi/4}2\sec x\,\mathrm dx=2\ln(\sqrt2+1)$
$I_2=\displaystyle\int_0^2\ln x\,\mathrm dx=2(\ln2-1)$
$I_3=\displaystyle\lim_{a\to\infty}\int_{-a}^a\frac{\mathrm dx}{x^2+1}=\pi$

Now,

$\sqrt2\approx1.4\implies \sqrt2+1\approx2.4$
$\implies \left\lceil I_1\right\rceil=2$

$e$
$\implies\left\lfloor I_2\right\rfloor=-1$

$\pi\approx3.14\implies\left\lceil I_3\right\rceil=4$

So the given equation reduces to

$\displaystyle\sum_{k=-1}^2\frac{\mathrm d}{\mathrm dx}x^{k+2}=1-4!$
$\dfrac{\mathrm dx}{\mathrm dx}+\dfrac{\mathrm dx^2}{\mathrm dx}+\dfrac{\mathrm dx^3}{\mathrm dx}+\dfrac{\mathrm dx^4}{\mathrm dx}=-23$
$4x^3+3x^2+2x+24=0$

a fairly standard cubic. Incidentally, when $x=-2$ , the LHS reduces to 0, so $x+2$ is a factor of the cubic. You can find the remaining two solutions easily with the quadratic formula.

5 0

2 years ago