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3 years ago

How much people can eat twice of 40 degree pizza wedges

Mathematics

2 answers:

babunello [35]3 years ago

8 0

20 beacause 20 x 2 = 40 and twice as many means 2 x ....

Gennadij [26K]3 years ago

3 0

20 because you make 40 half and it is 20,or you multiply something like 2✖️20 is equal to 40!!

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Elsa earns 5140 for working 20 hours

lys-0071 [83]

Answer:

$2.57

Step-by-step explanation:

divide 5,140 by 20,

5,140/20=257=2.57

7 0

2 years ago

What is the area of the square?

NISA [10]

The area is 4 square meters, because 2x2 would be 4

7 0

3 years ago

Allison was left an inheritance from her great aunt on her 12 birthday. However, she , she not able to touch it until she is 21

olya-2409 [2.1K]

Answer:

18% intrest

Step-by-step explanation:

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2 years ago

Select the correct answer from each drop-down menu.

vazorg [7]

Answer:

Step 1: Distribute $-2$ to $5x$ and $8$

Step 2: Subtract from both sides of the equation $6x$

Step 3: Add to both sides of the equation $16$

Step 4: Divide both sides of the equation by $-16$

Step-by-step explanation:

Step 1: Apply the Distributive Property. Then you must distribute $-2$ to $5x$ and $8$

Then:

$-2(5x+8)=14+6x\\\\-10x-16=14+6x$

Step 2: You must apply the Subtraction property of Equality and subtract $6x$ from both sides of the equation. Then:

$-10x-16-6x=14+6x-6x\\\\-16x-16=14$

Step 3: You must apply the Addition property of Equality and add $16$ to both sides of the equation. Then:

$-16x-16+16=14+16\\\\-16x=30$

Step 4: You must apply the Division property of Equality and divide both sides by $-16$ . Then:

$\frac{-16x}{-16}=\frac{30}{-16}\\\\x=\frac{30}{-16}\\\\x=-\frac{15}{8}$

6 0

3 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

2 years ago