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3 years ago

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

$260-68=192\\192=192$

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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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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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$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$

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What is the area of the parllelogtam shown below

Umnica [9.8K]

<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

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