Express 4 cm as a fraction of 2 m

Please help me i need to finish this i’m tired (please use grid if you can) asap!!!

jasenka [17]

The linear equation is y = 4x - 4

And the graph of the linear equation can be seen in the image below.

<h3>How to graph the last line?</h3>

It seems that you already are good at graphing, so I will try to explain how to find the equation more in detail.

Remember that a general linear equation is written as:

y = a*x + b

Where a is the slope and b is the y-intercept.

In this case, we know that the y-intercept is -4, then b = -4, replacing that we get:

y = a*x - 4

Now we also can see that this line passes through the point (2, 4), this means that if we evaluate in x = 2, the outcome should be y = 4, replacing that we get:

4 = a*2 - 4

4 + 4 = a*2

8 = a*2

8/2 = a = 4

Then the slope is 4, and the linear equation is:

y = 4x - 4

The graph is below.

If you want to learn more about linear functions:

brainly.com/question/4025726

#SPJ1

8 0

1 year ago

A university employs a total of 560 teaching assistants and research assistants. There are three times as many teaching

Scrat [10]

Answer:

170

Step-by-step explanation:

There are three times as many teaching assistants as research assistants. Find the number of research assistants employed by the university. Hence the number of research assistants is 170.

8 0

2 years ago

What is the quotient when 4x3 + 2x + 7 is divided by x + 3?

Arte-miy333 [17]

Answer:

The quotient of this division is $(4x^2 -12x + 38)$ . The remainder here would be $-26$ .

Step-by-step explanation:

The numerator $4x^3 + 2x + 7$ is a polynomial about $x$ with degree $3$ .

The divisor $x + 3$ is a polynomial, also about $x$ , but with degree $1$ .

By the division algorithm, the quotient should be of degree $3 - 1 = 2$ , while the remainder shall be of degree $1 - 1 = 0$ (i.e., the remainder would be a constant.) Let the quotient be $a\,x^2 + b\, x + c$ with coefficients $a$ , $b$ , and $c$ .

$4x^3 + 2x + 7 = \left(a\,x^2 + b\, x + c\right)(x + 3)$ .

Start by finding the first coefficient of the quotient.

The degree-three term on the left-hand side is $4 x^3$ . On the right-hand side, that would be $a\, x^3$ . Hence $a = 4$ .

Now, given that $a = 4$ , rewrite the right-hand side:

$\begin{aligned}&\left(4\,x^2 + b\, x + c\right)(x + 3) \cr =& \left(4x^2 + (b\, x + c)\right)(x + 3) \cr =& 4x^2(x + 3) + (bx + c)(x + 3) \cr =& 4x^3 + 12x^2 + (bx + c)(x + 3)\end{aligned}$ .

Hence:

$4x^3 + 2x + 7 = 4x^3 + 12x^2 + (b\,x + c)(x + 3)$

Subtract $\left(4x^3 + 12x^2\right$ from both sides of the equation:

$-12x^2 + 2x + 7 = (b\,x + c)(x + 3)$ .

The term with a degree of two on the left-hand side has coefficient $(-12)$ . Since the only term on the right hand side with degree two would have coefficient $b$ , $b = -12$ .

Again, rewrite the right-hand side:

$\begin{aligned}&\left(-12 x + c\right)(x + 3) \cr =& \left(-12 x+ c\right)(x + 3) \cr =& (-12x)(x + 3) + c(x + 3) \cr =& -12x^2 -36x + (bx + c)(x + 3)\end{aligned}$ .