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jeka94
3 years ago
12

Which of the following options have the same value as 96% of 25?

Mathematics
1 answer:
Masja [62]3 years ago
5 0

Answer:

A and B

Step-by-step explanation:

You might be interested in
In the diagram, the length of segment QV is 15 units.
Andrews [41]

Answer:

The answer is 15 units.

Step-by-step explanation:

The reason it is 15 units is because  We know that QV = 15 and TQ = TS, QV = SV;4 x - 1 = 154 x = 15 + 14 x = 16x = 16 : 4 = 4 henceforth the answer being 15 units

6 0
3 years ago
Read 2 more answers
What is u = kx + ух, for x
Ivahew [28]

Answer:

<h2>x=  \frac{u}{k + y}</h2>

Step-by-step explanation:

u = kx + ух

First of all factorize x out at the right side of the equation

That's

u = x(k + y)

Divide both sides by ( k + y) to make x stand alone

That's

<h3>\frac{x(k + y)}{k + y}  =  \frac{u}{k + y}</h3>

We have the final answer as

<h3>x =  \frac{u}{k + y}</h3>

Hope this helps you

6 0
3 years ago
Pls help soon:)
Tanzania [10]
Correct answers are:
(1) <span>28, 141 known cases
(2) 79913.71 known cases after six weeks (round off according to the options given)
(3) After approx. 9 weeks (9.0142 in decimal)

Explanations:
(1) Put x = 0 in given equation
</span><span>y= 28, 141 (1.19)^x
</span><span>y= 28, 141 (1.19)^(0)
</span>y= 28, 141


(2) Put x = 6 in the given equation:
<span>y= 28, 141 (1.19)^x
</span><span>y= 28, 141 (1.19)^(6)
</span>y= 79913.71

(3) Since
y= 28, 141 (1.19)^x
And y = <span>135,000

</span>135,000 = 28, 141 (1.19)^x
135,000/28, 141 = (1.19)^x

taking "ln" on both sides:
ln(4.797) = ln(1.19)^x

ln(4.797) = xln(1.19)
x = 9.0142 (in weeks)

5 0
3 years ago
Please help, I will give brainliest to best answer
tester [92]

Answer:

060.28

Step-by-step explanation:

3 can't go into 1, but it goes into 18 six times with 0 left over. 3 goes into 0 zero times. 3 goes into 5 one time with 2 left over. 3 goes into 26 (because you pull the 6 down) 8 times with 2 left over.

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
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