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3 years ago

Is this a function or not

Mathematics

1 answer:

tatuchka [14]3 years ago

7 0

Answer: neither are functions

Step-by-step explanation:

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The west room of the White House has a perimeter of 180 feet. If the width is 50 feet, what is the length of the West room?

NeTakaya

Answer:

l = 40

Step-by-step explanation:

Since Perimeter comes from P = 2l + 2w so 180 = 2l + 2(50)

180 = 2l + 100

180-100 = 2l

80 = 2l

so l = 40

4 0

3 years ago

Factor by grouping x^2-6x+5x-30

Gnoma [55]

x^2- 6x+5x-30
=x^2 - x -30
= (x - 6)(x + 5)

5 0

4 years ago

Linear or nonlinear? y = 2 with an exponent of two minus one

ELEN [110]

Answer:

it is a linear please mark me as brainliest

8 0

3 years ago

Consider the graph of the equation y = 5. Which statements are true? Check all that apply. The graph of y = 5 is a vertical line

lianna [129]

Hello!

y = 5 is a horizontal line whose y-values are all 5. Given this, the true statements are -

- The graph of y = 5 is a horizontal line
- The point (-4, 5) lies on the line

I hope this helps you!

5 0

3 years ago

Read 2 more answers

An=5-6n fifth term and sum of first three terms

Korolek [52]

Answer:

5th term is -25

and sum of first three terms is -21

Step-by-step explanation:

We are given the sequence:

$\displaystyle \large{a_n = 5 - 6n}$

To find 5th term, substitute n = 5.

$\displaystyle \large{a_5 = 5 - 6(5)} \\ \displaystyle \large{a_5 = 5 - 30} \\ \displaystyle \large{a_5 = - 25}$

Therefore, fifth term is 25.

Next, to find the sum of first three terms, we will introduce sigma.

$\displaystyle \large{a_1 + a_2 + a_3 + ... + a_n = \sum_{k = 1}^{n} a_k}$

Our ak is 5-6k

Since we want to find sum of first three terms:-

$\displaystyle \large{ \sum_{k = 1}^{3}( 5 - 6k)}$

Expand Sigma in.

$\displaystyle \large{ \sum_{k = 1}^{3} 5 + \sum_{k = 1}^{3}- 6k}$

Property of Summation

$\displaystyle \large{ \sum_{k = 1}^{n} m = m \times n \: \: \: \sf{(m \: \: is \: \: constant})} \\ \displaystyle \large{\sum_{k = 1}^{n}(a_k + b_k) = \sum_{k = 1}^{n}a_k + \sum_{k = 1}^{n}b_k} \\ \displaystyle \large{\sum_{k = 1}^{n}ma_k =m\sum_{k = 1}^{n} a_k \: \: \: \sf{(m \: \: is \: \: constant})}$

Therefore:-

$\displaystyle \large{ \sum_{k = 1}^{3} 5 + \sum_{k = 1}^{3}- 6k} \\ \displaystyle \large{ (5 \times 3) - 6\sum_{k = 1}^{3}k} \\ \displaystyle \large{ 15 - 6\sum_{k = 1}^{3}k}$

Summation Formula

$\displaystyle \large{ \sum_{k = 1}^{n}k = \frac{1}{2} n(n + 1) }$

Thus:-

$\displaystyle \large{ 15 - 6\sum_{k = 1}^{3}k} \\ \displaystyle \large{ 15 - 6( \frac{1}{2}(3)(3 + 1) } \\ \displaystyle \large{ 15 - 6( \frac{1}{2}(3)(4)) } \\\displaystyle \large{ 15 - 6(3)(2 )} \\\displaystyle \large{ 15 - 6(6)} \\ \displaystyle \large{ 15 -36 = - 21 }$

7 0

3 years ago