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sertanlavr [38]
3 years ago
14

5. Solve each equation by systematic trial.

Mathematics
1 answer:
Basile [38]3 years ago
3 0

Answer:

a) x = 260

b) g = 218

c) z = 30

d) h = 5

Step-by-step explanation:

a) x-68=192

Predict a number; let's say x = 225

225-68=192\\157=192

Too low; try with a higher number; let's say x = 270

270-68=192\\202=192

Too high, we're closer.

x = 260

260-68=192\\192=192

--------------------------------------------------------------------------------------

b) g+33=251

let's say g = 230

230+33=251\\263=251

Too high. Try g = 221

221+33=251\\254=251

We're almost there. Let's try with g = 218

218+33=251\\251=251

----------------------------------------------------------------------------------------

c) 3z-16=74

Let's try with z = 15

3(15)-16=74\\45-16=74\\29=74

Too low; try a higher number. z = 30

3(30)-16=74\\90-16=74\\74=74

--------------------------------------------------------------------------------------------

d) 15h+9=84

Let's try with h = 6

15(6)+9=84\\90+9=84\\99=84

Too high; let's try with h = 5

15(5)+9=84\\75+9=84\\84=84

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D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x
MA_775_DIABLO [31]
The solution depends on the value of k. To make things simple, assume k>0. The homogeneous part of the equation is

\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0

and has characteristic equation

r^2-16k=0\implies r=\pm4\sqrt k

which admits the characteristic solution y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}.

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be y_p=ae^{4x}+be^x. Then

\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x

So you have

16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x
(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x

This means

16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}
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and so the general solution would be

y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x
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What is the area of the parllelogtam shown below
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<h3>Answer: 16 square units</h3>

Let x be the height of the parallelogram. Right now it's unknown, but we can solve for it using the pythagorean theorem. Focus on the right triangle. It has legs a = 3 and b = x, with hypotenuse c = 5

a^2 + b^2 = c^2

3^2 + x^2 = 5^2

9 + x^2 = 25

x^2 = 25-9

x^2 = 16

x = sqrt(16)

x = 4

This is a 3-4-5 right triangle.

The height of the parallelogram is 4 units.

We have enough info to find the area of the parallelogram

Area of parallelogram = base*height

Area of parallelogram = 4*4

Area of parallelogram = 16 square units

Coincidentally, the base and height are the same, which isn't always going to be the case. The base is visually shown as the '4' in the diagram. The height is the dashed line, which also happens to be 4 units long.

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