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3 years ago

Help please I don’t understand:(

Mathematics

1 answer:

WITCHER [35]3 years ago

7 0

Here they are matched up not all of the lol
Have matches Hope this helps

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Andrew earned $125.00 last year on his investment of $10,000.00

Artemon [7]

Answer:

1,25%

Step-by-step explanation:

Aby policzyć stopę procentową należy wykorzystać podzielić otrzymany zarobek (125$) przez kwotę zainwestowaną (10.000$) a następnie pomnożyć przez 100, aby otrzymać wartość wyrażoną w procentach.

I tak mamy

$\frac{125}{10000}$ x 100 [%] = 1,25%

7 0

3 years ago

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I really need help for this geometry question, I would appreciate any help thank you.

Lynna [10]

Answer:

Step-by-step explanation:

In a 30 60 90 triangle the shortest side is considered to be half the hypotenuse. So in this case, 10 is the hypotenuse and 10/2 = 5.

7 0

3 years ago

Which describes the graph of y= -(x + 6)² + 6?

Eddi Din [679]

Answer:

Step-by-step explanation:

Since x has a negative coefficient, the graph opens downwards meaning that it has a maximum.

The vertex formula is y = a(x - h)^2 + k so the coordinate must be (-6, 6).

4 0

2 years ago

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Can someone help me with this

Alborosie

3 0

3 years ago

3y''-6y'+6y=e*x sexcx

Simora [160]

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

$3r^2-6r+6=0\iff r^2-2r+2=0$

which has roots at $r=1\pm i$ . This admits the two fundamental solutions

$y_1=e^x\cos x$
$y_2=e^x\sin x$

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

$u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx$
$u_2=\dfrac13x$

Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

7 0

3 years ago