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levacccp [35]
3 years ago
8

Hernando ate StartFraction 2 Over 8 EndFraction of a pizza for a dinner. He gave his 6 friends the rest of the pizza and told th

em to share it equally. To determine the fraction of the whole pizza that each of his friends received, Hernando performed the calculations below. StartFraction 6 Over 8 EndFraction divided by 6 = StartFraction (6 divided by 6) Over 8 EndFraction = one-eighth He concluded that each person received One-eighth of the whole pizza. Are Hernando's calculations correct? Explain why or why not. Yes, they are correct. Hernando divided 6 by 6. Yes, they are correct. Hernando divided Two-eighths by 6. No, they are not correct. Hernando should have multiplied by 6, not divided by 6. No, they are not correct. Hernando should have divided 8 by 6, not 6 by 6.
Mathematics
2 answers:
Akimi4 [234]3 years ago
8 0

Answer:

A Yes, they are correct. Hernando divided 6 by 6. on edge

Step-by-step explanation:

hope this helps:)

ozzi3 years ago
3 0

Answer:

Yes, they are correct. Hernando divided 6 by 6.

Step-by-step explanation:

Given that:

Hernando ate 2/8 of a pizza for dinner.

Let the pizza be x. Then:

Hernando ate \dfrac{2}{8} \ \ o f  \ \ x

Then he gave the rest of the pizza to his friends

Hernando now carried out the  following calculations to support his activities;

\dfrac{6}{8} \div 6= \dfrac{6 \div 6}{8}= \dfrac{1}{8}

\dfrac{6}{8} \times \dfrac{1}{6}= \dfrac{6 \div 6}{8}= \dfrac{1}{8}

From the above calculation; we will see that Hernando divided 6 by 6 in both cases to get \dfrac{1}{8}.

Thus, the correct answer is:

Yes, they are correct. Hernando divided 6 by 6.

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Write the limit as a definite integral on the interval [a, b], where ci is any point in the ith subinterval. Limit Interval lim
geniusboy [140]

Answer:

The corresponding definite integral may be written as

\int_a^b \mathrm{(4x + 11)}\,\mathrm{d}x

The answer of the above definite integral is

\int_a^b \mathrm{(4x + 11)}\,\mathrm{d}x = 98

Step-by-step explanation:

The given limit interval is

\lim_{||\Delta|| \to 0}  \sum\limits_{i=1}^n (4c_i + 11) \Delta x_i

[a, b] =  [-8, 6]

The corresponding definite integral may be written as

\int_a^b \mathrm{(4x + 11)}\,\mathrm{d}x

\int_{-8}^6 \mathrm{(4x + 11)}\,\mathrm{d}x

Bonus:

The definite integral may be solved as

\int_{-8}^6 \mathrm{(4x + 11)}\,\mathrm{d}x \\\\\frac{4x^2}{2} + 11x \left \|{b=6} \atop {a=-8}} \right. \\\\2x^2 + 11x \left \|{b=6} \atop {a=-8}} \right. \\\\ 2(6^2 -(-8)^2 ) + 11(6 - (-8) \\\\2(36 - 64 ) + 11(6 + 8) \\\\2(-28 ) + 11(14) \\\\-56 +154 \\\\98

Therefore, the answer to the integral is

\int_a^b \mathrm{(4x + 11)}\,\mathrm{d}x  = 98

5 0
3 years ago
Help!!!!!!!!!!!!!!!!!!!!!!!!!!
morpeh [17]

Answer:

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8 0
3 years ago
Write the values that make the denominators zero. then solve the equation?
Olin [163]

Answer:

  denominators are zero for x=1

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Step-by-step explanation:

We suspect your equation is supposed to be ...

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The denominators will be zero for x=1.

This version of the equation has no solution. It simplifies to ...

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<em>Alternative interpretation</em>

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4 0
3 years ago
What is the sum of the interior angles of the polygon pictured below?
kkurt [141]

3420°

explanation:

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8 0
3 years ago
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Dafna1 [17]
6.11

see- .8 .9 .10 so its .11
6 0
3 years ago
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