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3 years ago

so here's my problem from pearson realize lesson 4-8 grade 7 math book. can someone help me int this question

Mathematics

1 answer:

Allushta [10]3 years ago

6 0

a) 0.92v or 23v/25 or 92%v

b)Oringaly, there was 100%. 8% was token away, so 100%-8%=92%

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d1i1m1o1n [39]

Step-by-step explanation:

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3 years ago

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Geri believed the answer to one of her homework problems was 2 cubed. After checking her work, she realized the correct answer w

TiliK225 [7]

Answer:

The correct answer is 64 times Geri’s original answer

Step-by-step explanation:

Given

Gari's Answer: 2³

Correct Answer: 2⁹

Required:

Relationship between both answers

Let Gari's answer be represented by G and the Correct answer be represented by C; Such that

G = 2³

C = 2⁹

Divide C by G

$\frac{C}{G} = \frac{2^9}{2^3}$

From law of indices, $\frac{a^m}{a^n} = a^{m-n}$

Applying this law, we have

$\frac{C}{G} = 2^{9-3}$

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$\frac{C}{G} = 2*2*2*2*2*2$

$\frac{C}{G} = 64$

Multiply both sides by G

$G * \frac{C}{G} = 64 * G$

$C = 64G$

Hence, we can conclude that the correct answer is 64 times Geri’s original answer

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3 years ago

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Effectus [21]

Answer:

1) True 2) False

Step-by-step explanation:

1) Given $\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$

To verify that the above equality is true or false:

Now find $\sum\limits_{k=0}^8\frac{1}{k+3}$

Expanding the summation we get

$\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{0+3}+\frac{1}{1+3}+\frac{1}{2+3}+\frac{1}{3+3}+\frac{1}{4+3}+\frac{1}{5+3}+\frac{1}{6+3}+\frac{1}{7+3}+\frac{1}{8+3}$ $\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$