All categories

2 years ago

Find the equation of the linear function represented by the table below in slope-

Mathematics

2 answers:

kramer2 years ago

7 0

Answer:

y=3x+7

Step-by-step explanation:

take two points, (1,13) (2,19)

subtract using slope formula

19-13 2-1= 6/1 -> 6-> 3

find your y intercept (b) by using the first term in the 7 row (7)

Rzqust [24]2 years ago

4 0

Answer:

y = 3x +

Step-by-step explanation:

The equation of the line is given by:

----------------------------

The equation of a line, in slope-intercept formula, has the following format:

In which

m is the slope, that is, how much y changes when x changes by 1.

b is the y-intercept, that is, the value of y when x = 0.

----------------------------

Given two points (x,y), the slope is the change in y divided by the change in x.

We have points (-1, -1) and (3,11). Thus:

Then

----------------------------

Point (-1,-1) means that when , and we use this to find b.

Thus

A similar problem is given at brainly.com/question/13251894

Still stuck? Get 1-on-1 help from an expert tutor now.

You might be interested in

What is the remainder of (3x^4+2x^3-x^2+2x-14)/ (x+2)

Allisa [31]

Hi,

I assume the forward slash meant it as a fraction VS it being a division symbol. To answer your question, the remainder is 10.

Hope this helps.

4 0

3 years ago

Complete the number sentence to show an example of the Distributive Property. 3×6=

Evgesh-ka [11]

If you want to know how much is 3x6 the answer is 18

7 0

3 years ago

Read 2 more answers

Please help me to prove this!

Sophie [7]

Answer: see proof below

<u>Step-by-step explanation:</u>

Given: A + B = C → A = C - B

→ B = C - A

Use the Double Angle Identity: cos 2A = 2 cos² A - 1

→ (cos 2A + 1)/2 = cos² A

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · 2 cos [(A - B)/2]

Use Even/Odd Identity: cos (-A) = cos (A)

<u>Proof LHS → RHS:</u>

LHS: cos² A + cos² B + cos² C

$\text{Double Angle:}\qquad \dfrac{\cos 2A+1}{2}+\dfrac{\cos 2B+1}{2}+\cos^2 C\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}\bigg(2+\cos 2A+\cos 2B\bigg)+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\dfrac{1}{2}\bigg(\cos 2A+\cos 2B\bigg)+\cos^2 C$

$\text{Sum to Product:}\quad 1+\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\cos (A+B)\cdot \cos (A-B)+\cos^2 C$

$\text{Given:}\qquad \qquad 1+\cos C\cdot \cos (A-B)+\cos^2C$

$\text{Factor:}\qquad \qquad 1+\cos C[\cos (A-B)+\cos C]$

$\text{Sum to Product:}\quad 1+\cos C\bigg[2\cos \bigg(\dfrac{A-B+C}{2}\bigg)\cdot \cos \bigg(\dfrac{A-B-C}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+(C-B)}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-(C-A)}{2}\bigg)$

$\text{Given:}\qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+A}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-B}{2}\bigg)\\\\\\.\qquad \qquad \qquad =1+2\cos C \cdot \cos A\cdot \cos (-B)$