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4 years ago

Which statement describes the term salary

1 answer:

3 0

Answer: Best described as a fixed amount of income that is paid to an someone weekly, every other week, or on a monthly basis. it could also be described as annual earnings paid out over a year.

Step-by-step explanation:

You might be interested in

What is the distance around a triangle with the sides measuring 2⅛, 3½ and 2½?

Marta_Voda [28]

So when you have data given this kind of form... you should convert to decimal first.
So that would be:
a ≈ 2.12
b = 3.5
c = 2.5
Now just place the simple perimeter formula, and sum all sides.

7 0

3 years ago

If a parachutist jumped from an airplane flying at an altitude of 10,000 feet and fell at an average free fall speed of 158 feet

scZoUnD [109]

The answer is -2,686

let me tell you why !

the answer would be negative, since the parachutist is descending. so 158 × 17 = 2,686

make that negative.

you get -2,686 !!!! hope that helps :)))

7 0

3 years ago

Use the figure below to complete the following problem Given : R,S,T are midpoints of AC, AB, and CB.

Nikolay [14]

Answer:

answer=RT

Step-by-step explanation:

because

3 0

3 years ago

Read 2 more answers

Find AB using the given matrices.

solniwko [45]

The entry in row 1, column 1 of $\mathbf{AB}$ is

$\begin{bmatrix}2&4\end{bmatrix}\begin{bmatrix}5\\6\end{bmatrix}=2\cdot5+4\cdot6=34$

(i.e. the dot product of row 1 of $\mathbf A$ and column 1 of $\mathbf B$ )

The entry in row 1, column 2 of $\mathbf{AB}$ is

$\begin{bmatrix}2&4\end{bmatrix}\begin{bmatrix}-9\\-4\end{bmatrix}=2\cdot(-9)+4\cdot(-4)=-34$

The second option is the correct answer.

5 0

3 years ago

can someone show me how to find the general solution of the differential equations? really need to know how to do it for the upc

mariarad [96]

The first equation is linear:

$x\dfrac{\mathrm dy}{\mathrm dx}-y=x^2\sin x$

Divide through by $x^2$ to get

$\dfrac1x\dfrac{\mathrm dy}{\mathrm dx}-\dfrac1{x^2}y=\sin x$

and notice that the left hand side can be consolidated as a derivative of a product. After doing so, you can integrate both sides and solve for $y$ .

$\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1xy\right]=\sin x$
$\implies\dfrac1xy=\displaystyle\int\sin x\,\mathrm dx=-\cos x+C$
$\implies y=-x\cos x+Cx$

- - -

The second equation is also linear:

$x^2y'+x(x+2)y=e^x$

Multiply both sides by $e^x$ to get

$x^2e^xy'+x(x+2)e^xy=e^{2x}$

and recall that $(x^2e^x)'=2xe^x+x^2e^x=x(x+2)e^x$ , so we can write

$(x^2e^xy)'=e^{2x}$
$\implies x^2e^xy=\displaystyle\int e^{2x}\,\mathrm dx=\frac12e^{2x}+C$
$\implies y=\dfrac{e^x}{2x^2}+\dfrac C{x^2e^x}$

- - -

Yet another linear ODE:

$\cos x\dfrac{\mathrm dy}{\mathrm dx}+\sin x\,y=1$

Divide through by $\cos^2x$ , giving

$\dfrac1{\cos x}\dfrac{\mathrm dy}{\mathrm dx}+\dfrac{\sin x}{\cos^2x}y=\dfrac1{\cos^2x}$
$\sec x\dfrac{\mathrm dy}{\mathrm dx}+\sec x\tan x\,y=\sec^2x$
$\dfrac{\mathrm d}{\mathrm dx}[\sec x\,y]=\sec^2x$
$\implies\sec x\,y=\displaystyle\int\sec^2x\,\mathrm dx=\tan x+C$
$\implies y=\cos x\tan x+C\cos x$
$y=\sin x+C\cos x$

- - -

In case the steps where we multiply or divide through by a certain factor weren't clear enough, those steps follow from the procedure for finding an integrating factor. We start with the linear equation

$a(x)y'(x)+b(x)y(x)=c(x)$

then rewrite it as

$y'(x)=\dfrac{b(x)}{a(x)}y(x)=\dfrac{c(x)}{a(x)}\iff y'(x)+P(x)y(x)=Q(x)$

The integrating factor is a function $\mu(x)$ such that

$\mu(x)y'(x)+\mu(x)P(x)y(x)=(\mu(x)y(x))'$

which requires that

$\mu(x)P(x)=\mu'(x)$

This is a separable ODE, so solving for $\mu$ we have

$\mu(x)P(x)=\dfrac{\mathrm d\mu(x)}{\mathrm dx}\iff\dfrac{\mathrm d\mu(x)}{\mu(x)}=P(x)\,\mathrm dx$
$\implies\ln|\mu(x)|=\displaystyle\int P(x)\,\mathrm dx$
$\implies\mu(x)=\exp\left(\displaystyle\int P(x)\,\mathrm dx\right)$

and so on.

6 0

4 years ago