Write two expressions with a gcf of 8xy

10. The fraction 325/790 converted to a decimal and rounded to the nearest hundredths place is _____

Romashka [77]

Answer: 325/790 rounded to the nearest hundredths place is <em>0.41</em>

Step-by-step explanation:

if you divide 325/790 you get 0.411 and some other numbers. We only need to focus on these 3 numbers. 4 is the tenths place value and the first 1 is in the hundredths place value. Sense the second 1 (which is in the thousandths place value) is less than 5, we don't round the 1 in the hundredths place value up. Meaning your answer is 0.411.

Hopefully this makes sense.

5 0

3 years ago

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PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!!

wolverine [178]

Answer:

the green line

Step-by-step explanation:

its closest to the point

4 0

2 years ago

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In kilometers, the approximate distance to the earth's horizon from a point h meters above the surface can be determined by eval

Maslowich

In kilometers, the approximate distance to the earth's horizon from a point h meters above the surface can be determined by evaluating the expression

$d= \sqrt{12h}$

We are given the height h of a person from surface of sea level to be 350 m and we are to find the the distance to horizon d. Using the value in above expression we get:

$d= \sqrt{12*350}=64.81$

Therefore, the approximate distance to the horizon for the person will be 64.81 km

8 0

3 years ago

What is the effect on the graph of the function f(x) = |x| when f(x) is changed to f(x + 9)?

babymother [125]

The answer is A) shifts right by 9 units

3 0

3 years ago

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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

2 years ago