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3 years ago

Solve. 3(5z + 7) = 33

Mathematics

2 answers:

Alex_Xolod [135]3 years ago

5 0

Answer:

z = 4/5

Step-by-step explanation:

Ivenika [448]3 years ago

3 0

Answer:

z = 4/5

Step-by-step explanation:

3(5z + 7) = 33

Divide each side by 3

3/3(5z + 7) = 33/3

5z+7 = 11

Subtract 7 from each side

5z+7-7 = 11-7

5z = 4

Divide by 5

5z/5 = 4/5

z = 4/5

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Divide. Write in simplest form. 10 divided by 1 2/3 ?

adell [148]

Answer: 6

Step-by-step explanation: First rewrite 10 as 10/1 and 1 and 2/3 as 5/3.

Mixed numbers can be changed to improper fractions by multiplying the denominator by the whole number and then adding the numerator. We then put out numerator over our old denominator.

So we have 10/1 ÷ 5/3 or 10/1 × 3/5.

It's important to understand that dividing by a fraction is the same as multiplying by its reciprocal. In other words, we can change the division to multiplication and flip the second fraction.

Now multiplying across the numerators and across the denominators, we have 30/5. Notice however that 30/5 is not in lowest terms so we divide the numerator and the denominator by the greatest common factor of 30 and 5 which is 6 and we end up with 6.

Therefore, 10 ÷ 1 and 2/3 = 6.

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2 years ago

Estimate the area of each figure. each square represents 1 square foot.

maksim [4K]

I thinks it’s 25ft ^2

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2 years ago

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Please solve ASAP!!!!!!!!!

Ulleksa [173]

Answer:

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16. B

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Step-by-step explanation:

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2 years ago

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What is the sum of the first 51 consecutive odd positive integers?

Angelina_Jolie [31]

We call:

$a_{n}$ as the set of the first 51 consecutive odd positive integers, so:

 $a_{n} = \{1, 3, 5, 7, 9...\}$

Where:
$a_{1} = 1$
$a_{2} = 3$
$a_{3} = 5$
$a_{4} = 7$
$a_{5} = 9$
and so on.

In mathematics, a sequence of numbers, such that the difference between two consecutive terms is constant, is called Arithmetic Progression, so:

3-1 = 2
5-3 = 2
7-5 = 2
9-7 = 2 and so on.

Then, the common difference is 2, thus:

 $a_{n} = \{ a_{1} , a_{1} + d, a_{1} + d + d,..., a_{1} + (n-2)d+d\}$

Then:

 $a_{n} = a_{1} + (n-1)d$

So, we need to find the sum of the members of the finite series, which is called arithmetic series:

There is a formula for arithmetic series, namely:

 $S_{k} = ( \frac{a_{1} + a_{k}}{2} ).k$

Therefore, we need to find:
 $a_{k} = a_{51}$

Given that $a_{1} = 1$ , then:

$a_{n} = a_{1} + (n-1)d = 1 + (n-1)(2) = 2n-1$

Thus:
$a_{k} = a_{51} = 2(51)-1 = 101$

Lastly:

$S_{51} = ( \frac{1 + 101}{2} ).51 = 2601$

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vfiekz [6]

All of them are correct

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3 years ago

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