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
Answer:
<h2>

</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>

</h3>
We have the final answer as
<h3>

</h3>
Hope this helps you
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)
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.
The first equation is linear:

Divide through by

to get

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

.
![\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1xy\right]=\sin x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5B%5Cdfrac1xy%5Cright%5D%3D%5Csin%20x)


- - -
The second equation is also linear:

Multiply both sides by

to get

and recall that

, so we can write



- - -
Yet another linear ODE:

Divide through by

, giving


![\dfrac{\mathrm d}{\mathrm dx}[\sec x\,y]=\sec^2x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5B%5Csec%20x%5C%2Cy%5D%3D%5Csec%5E2x)



- - -
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

then rewrite it as

The integrating factor is a function

such that

which requires that

This is a separable ODE, so solving for

we have



and so on.