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3 years ago

What us 8171.28 in 1 d.p

Mathematics

1 answer:

expeople1 [14]3 years ago

6 0

Answer:

Step-by-step explanation:

8171.28

Sig Figs

8171.28

Decimals

8171.28

Scientific Notation

8.17128 × 103

E-Notation

8.17128e+3

Words

eight thousand one hundred seventy-one point two eight

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1,000 * 32 = 32,000

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2 years ago

What is the solution to the inequality below?

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C. x < 25 and x ≥ 0

Step-by-step explanation:

Fastest and easiest way to do this is to graph the inequality and find out the lines.

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3 years ago

What is the distance on the unit circle between successive fourth roots of StartFraction StartRoot 3 EndRoot Over 2 EndFraction

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3 years ago

Read 2 more answers

1. Solve for x. Then find the angle measures of the<br>triangle?<br>(4x)<br>(6x+55)<br>(9x+30)

nalin [4]

Answer:

x = 5.

First angle = 20°, Second angle = 85°, Third angle = 75°

Step-by-step explanation:

The first angle of the triangle = (4 x)

The second angle of the triangle = (6x + 55)

The third angle of the triangle = (9x + 30)

By ANGLE SUM PROPERTY of a triangle:

First angle + Second angle + Third angle = 180°

⇒ (4 x) + (6x+55) + (9x+30) = 180°

or, (4x + 6x + 9x) + ( 55 + 30) = 180°

or, 19x = 180 - 85

or, 19 x = 95 ⇒ x = 95 /19 = 5

or, x = 5

Hence, first angle of the triangle = (4 x) = 4 x 5 = 20°

Second angle = ( 6x + 30) = 6(5) + 55 = 85°

Third angle = (9x + 30 = 9(5) + 30 = 75°

8 0

3 years ago

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

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

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

8 0

3 years ago

What us 8171.28 in 1 d.p​

What us 8171.28 in 1 d.p