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2 years ago

I need help with this

Mathematics

1 answer:

Mademuasel [1]2 years ago

5 0

Answer:

30/90 = 1/3

60/120 = 1/2

7/49 = 1/7

if you're doing mixed numbers:

30/90 = 3

60/120= 0 1/5

7/49= I couldn't tell ya that one exactly.

Step-by-step explanation:

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Which choice shows the coordinates of B’ if the trapezoid is reflected across the x-axis?

Strike441 [17]

For this case we have that the original point is given by:
B = (7, 2)
As the point is reflected through the x axis, then we have the following transformation:
(x, y) --------------> (x, -y) -------------> (x ', y')
Applying the transformation to the original ordered pair we have:
(7, 2) --------------> (7, - (2)) -------------> (7, -2)
Answer:
Point B 'is given by:
B '= (7, -2)

8 0

2 years ago

Read 2 more answers

Please help with this

timurjin [86]

Answer:

Hey mate.....

Step-by-step explanation:

This is ur answer.....

Step 1 :-

Step 2 :-

Hope it helps!

Follow me! ;)

4 0

2 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

3 years ago

Match these items. Match the items in the left column to the items in the right column

Anni [7]

Answer:

sorry I can't solve it like this where is the right column

4 0

3 years ago

Reflect (-6, -7) in the x-axis thanks:)

GenaCL600 [577]

(-6,7) since it’s the x axis you just change the sign of the y and keep x the same

3 0

2 years ago

I need help with this ​

I need help with this