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3 years ago

Consider the expressions 12 + 4y and 4( y + 3 )what is the value of each expression when y = 5

Mathematics

1 answer:

Neko [114]3 years ago

4 0

Answer:

Both expressions equals to 32.

Step-by-step explanation:

You just plug in 5 for each expression: 12+4(5)=32, 4(5+3)=32.

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What is the surface area of this right circular cone?

docker41 [41]

Answer:

○ B) 624π mm.²

Step-by-step explanation:

$\pi r(r + \sqrt{ {r}^{2} + {h}^{2}}) = S. A. \\ \\ \pi (12)(12 + \sqrt{{12}^{2} + {40}^{2}}) \\ \\ 12\pi(12 + \sqrt{1744}) = 12\pi(12 + 38) = 12\pi(50) = 600\pi \\ \\ 600\pi ≈ 624\pi$

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3 0

3 years ago

The ratio of cats to dogs in the shelter is 2:3. If there are 8 cats, how many dogs are there?

maxonik [38]

12, for every 2 cats there are three dogs. 8/2 is four. you multiply 3 times 4 to get 12

4 0

1 year ago

Read 2 more answers

Suppose you are representing the employees at a large corporation during contract negotiations. You have a list of the salaries

evablogger [386]

Answer:

Median

Step-by-step explanation:

Median is ascertained by arranging a set of data in either ascending or descending order and picking the number that appears In the middle. The measure of center to use for this test is the median because it will represent the employees, it helps to arrange their income accordingly and the central item is determined.

7 0

3 years ago

A = 1011 + 337 + 337/2 +1011/10 + 337/5 + ... + 1/2021

egoroff_w [7]

The sum of the given series can be found by simplification of the number

of terms in the series.

A is approximately <u>2020.022</u>

Reasons:

The given sequence is presented as follows;

A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021

Therefore;

$\displaystyle A = \mathbf{1011 + \frac{1011}{3} + \frac{1011}{6} + \frac{1011}{10} + \frac{1011}{15} + ...+\frac{1}{2021}}$

The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;

$\displaystyle a_{n+1} = \mathbf{\frac{n^2 + 3 \cdot n + 2}{2}}$

Therefore, for the last term we have;

$\displaystyle 2043231= \frac{n^2 + 3 \cdot n + 2}{2}$

2 × 2043231 = n² + 3·n + 2

Which gives;

n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0

Which gives, the number of terms, n = 2020

$\displaystyle \frac{A}{2} = \mathbf{ 1011 \cdot \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460} \right)}$

$\displaystyle \frac{A}{2} = 1011 \cdot \left(1 - \frac{1}{2} +\frac{1}{2} - \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022} \right)$