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2 years ago

Given the area of a square is 225 ft.² what is the length of each side

Mathematics

1 answer:

Vinvika [58]2 years ago

8 0

Answer:

15ft

Step-by-step explanation:

the square root of 225 is 15, so the answer is 15ft

15×15=225

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15 feet 10 inches is how many inches in all

Lady bird [3.3K]

Answer:

190 inches

Step-by-step explanation:

(15x12)+10=190

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2 years ago

Pls help i need this done as quick as possible

Anvisha [2.4K]

Answer:

Infinite Solutions.

Step-by-step explanation:

$m+7=3+m+4$

Subtract m from both sides:

$m+7-m=3+m+4-m$

$7=3+4$

$7=7$

Both sides are equal.

This means that the value of m does not affect the end result of 7 = 7, which is true. Thus, there are infinite solutions as any value of m will still make the equation true.

4 0

2 years ago

The question is in the picture, PLEASE HELPPP!

Mumz [18]

Answer:

1=c

Step-by-step explanation:

1=c

2=d

3=e

4=a

5=b

I know 1 and 2 are correct, but I am not sure about the others

I'm not too sure how to explain this, sorry

8 0

2 years ago

Find the volume of the cone. H=9cm r=6cm

never [62]

Answer:

the volume of the cone is 339.29

Step-by-step explanation:

3 0

2 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

2 years ago