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3 years ago

Is $15 per hour a unit rate?

Mathematics

1 answer:

Blababa [14]3 years ago

5 0

Yes. have a great day/night!

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How would I do this Need help

Annette [7]

10^2= 2^2 x a^2

a^2= 100-4

a= √96

a= 9.79..

a= 9.8

7 0

3 years ago

Read 2 more answers

Solve the equation for x. 27(x - 108) = 54

Sidana [21]

X = 110

110-108 = 2

27*2=54

4 0

3 years ago

Read 2 more answers

Solve. -1/3b = 9 A.-3 B.3 C.-27 D.27

STALIN [3.7K]

Ok so lets plug in a first -1/3(-3) would equal 9 a is the accurate answer

3 0

3 years ago

Read 2 more answers

Please and thank you to whoever helps

Zepler [3.9K]

Answer:

1131.0 m³

Step-by-step explanation:

Using the formula for the volume of a cylinder, fill in the appropriate values and do the arithmetic.

V = πr²h

Since this formula needs the radius of the cylinder and you are given the diameter, you must divide the diameter by 2 to get the radius:

r = d/2 = (12 m)/2 = 6 m

Then the volume is ...

V = π(6 m)²(10 m) = 360π m³ ≈ 1131.0 m³ . . . . . rounded to tenths

8 0

4 years ago

Use the Newton-Raphson method to find the root of the equation f(x) = In(3x) + 5x2, using an initial guess of x = 0.5 and a stop

xxMikexx [17]

Answer with explanation:

The equation which we have to solve by Newton-Raphson Method is,

f(x)=log (3 x) +5 x²

$f'(x)=\frac{1}{3x}+10 x$

Initial Guess =0.5

Formula to find Iteration by Newton-Raphson method

$x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}\\\\x_{1}=x_{0}-\frac{f(x_{0})}{f'(x_{0})}\\\\ x_{1}=0.5-\frac{\log(1.5)+1.25}{\frac{1}{1.5}+10 \times 0.5}\\\\x_{1}=0.5- \frac{0.1760+1.25}{0.67+5}\\\\x_{1}=0.5-\frac{1.426}{5.67}\\\\x_{1}=0.5-0.25149\\\\x_{1}=0.248$

$x_{2}=0.248-\frac{\log(0.744)+0.30752}{\frac{1}{0.744}+10 \times 0.248}\\\\x_{2}=0.248- \frac{-0.128+0.30752}{1.35+2.48}\\\\x_{2}=0.248-\frac{0.17952}{3.83}\\\\x_{2}=0.248-0.0468\\\\x_{2}=0.2012$

$x_{3}=0.2012-\frac{\log(0.6036)+0.2024072}{\frac{1}{0.6036}+10 \times 0.2012}\\\\x_{3}=0.2012- \frac{-0.2192+0.2025}{1.6567+2.012}\\\\x_{3}=0.2012-\frac{-0.0167}{3.6687}\\\\x_{3}=0.2012+0.0045\\\\x_{3}=0.2057$

$x_{4}=0.2057-\frac{\log(0.6171)+0.21156}{\frac{1}{0.6171}+10 \times 0.2057}\\\\x_{4}=0.2057- \frac{-0.2096+0.21156}{1.6204+2.057}\\\\x_{4}=0.2057-\frac{0.0019}{3.6774}\\\\x_{4}=0.2057-0.0005\\\\x_{4}=0.2052$

So, root of the equation =0.205 (Approx)

Approximate relative error

$=\frac{\text{Actual value}}{\text{Given Value}}\\\\=\frac{0.205}{0.5}\\\\=0.41$

Approximate relative error in terms of Percentage

=0.41 × 100

= 41 %

7 0

3 years ago