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3 years ago

Very sleepy plz help, been doing these all day... question in photo

Mathematics

1 answer:

zzz [600]3 years ago

6 0

Answer:

20 40 and 90 are correct

Step-by-step explanation:

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The fee to check luggage with anew arline is based on the weightof each bag. Brett paid $7.14 tocheck a 42-pound bag. Howmuch wi

rodikova [14]

$10.71

Explanation

Step 1

you can easily solve this by using a rule of three.

Let

x represents the money Eileen will pay if her bag weighs 63 pounds

the proportion is the same

then

$\frac{7.14}{42}=\frac{x}{63}$

Step 2

solve for x

$\begin{gathered} \frac{7.14}{42}=\frac{x}{63} \\ 63\cdot7.14=42x \\ x=\frac{63\cdot7.14}{42} \\ x=\frac{449.82}{42} \\ x=10.71 \\ \end{gathered}$

I hope this helps you

5 0

1 year ago

Help with word problem!

creativ13 [48]

2 1/6 divided by 7/10

13/6 / 7/10 =

13/6 * 10/7 =

130/42 =

3 4/42. or 3 2/21 hours... three and two twenty-firsts

7 0

3 years ago

At a pet store, Davina counted 12 parrots out of 20 birds. Which is an equivalent ratio of parrots to birds at the pet store?

rodikova [14]

D. 3 to 5 This is because if you simplify the ratio by dividing each number in the ratio by 4, therefore for every group of 5 birds 3 of them are parrots

5 0

3 years ago

Read 2 more answers

Carlo spent $107.60 on 8 books. If each book cost the same amount, how much did one book cost?

dlinn [17]

Answer:

$13.45

Step-by-step explanation:

Divide $107.60 by 8

4 0

3 years ago

Read 2 more answers

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