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2 years ago

What is the y-intercept of the line given by the equation y= 3x + 4?

Mathematics

2 answers:

DIA [1.3K]2 years ago

8 0

Answer:

the y intercept is 4

Step-by-step explanation:

the 4 is where the line starts so it would be (0,4)

Alik [6]2 years ago

8 0

First, notice that this equation is written in slope-intercept form which is more commonly known as y = mx + b form where the multiplier, m, or the coefficient of the ex term represents the slope of the line and the b or the constant term represents the y-intercept of the line.

I have attached slope-intercept form below.

Notice that below, the b is in place of the 4 in the equation.

So the y-intercept or the b is 4.

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A company offers four plans for customers who want to rent DVDs and video games. The cost of the plans can be modeled by a linea

Luba_88 [7]

To solve this, we are going to use the slope formula, and the point slope formula.
Slope formula: $m= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
where
$m$ is the slope.
$(x_{1},y_{1})$ are the coordinates of the first point.
$(x_{2},y_{2})$ are the coordinates of the first point.
Point-slope formula: $y-y_{1}=m(x-x_{1})$

Plan A. Coordinates of the first point: (1,14) (number of discs, cost). Coordinates of the second point: (2,17). Lets replace those values in our slope formula:
$m= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$m= \frac{17-14}{2-1}$
$m=3$
Now that we have the slope, we can use the point slope formula:
$y-y_{1}=m(x-x_{1})$
$y-14=3(x-1)$
$y-14=3x-3$
$y=3x+11$
Remember that in a linear equation of the form $y=mx+b$ , $m$ is the slope and $b$ is the y-intercept. In our model the y-intercept is the membership fee.
Since $b=11$ in our linear equation, we can conclude that the membership fee is $11

Plan B. Coordinates of the first point (1,12). Coordinates of the second point (2,16).
$m= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$m= \frac{16-12}{2-1}$
$m=4$
$y-y_{1}=m(x-x_{1})$
$y-12=4(x-1)$
$y-12=4x-4$
$y=4x+8$
Membership fee: $8

Plan C. Coordinates of the first point: (1,10). Coordinates of the second point: (2,15).
$m= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$m= \frac{15-10}{2-1}$
$m=5$
$y-y_{1}=m(x-x_{1})$
$y-10=5(x-1)$
$y-10=5x-5$
$y=5x+5$
Membership fee: $5

Part D. Coordinates of the first point: (1,12). Coordinates of the second point (2,21).
$m= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$m= \frac{21-12}{2-1}$
$m=9$
$y-y_{1}=m(x-x_{1})$
$y-12=9(x-1)$
$y-12=9x-9$
$y=9x+3$
Membership fee: $3

We can conclude that the plan that has the smallest one-time membership is Plan D.

6 0

2 years ago

Read 2 more answers

In the last year, U.S. car sales of all makes and models have dropped by 16%. Let's say that a car dealer sells 325 cars. If thi

IRINA_888 [86]

Answer:

c. 273 cars

Step-by-step explanation:

First you need to find what 16% of 325 is

325x.16=52

because the number sold dropped by 16% you will subtract 52 from 325

325-52=273

you can also calculate the answer by subtracting 100% by 16%

100%-16%=84%

now you can calculate 84% of 325

325x.84=273

6 0

2 years ago

Please do 3 & 4 and show work

Shalnov [3]

Answer:

3. A=B=70°; C=D=110°

4. A=90°; C=48°

Step-by-step explanation:

Same-side angles of a trapezoid add to 180°.

___

3. (2x) +(x+15) = 180

3x + 15 = 180 . . . . simplify

x +5 = 60 . . . . . . . divide by 3

x = 55

A=B=(x+15)° = (55+15)° = 70°

C=D=2x° = 2·55° = 110°

4. You are given the values of one of the angles on each side, so the other can be found by subtraction:

A=180°-D = 90°

C = 180° -132° = 48°

8 0

2 years ago

Please please help What is the dependent variable? A) height B )armspan

monitta

Height because dependent variable is the measurement

4 0

2 years ago

Suppose z equals f (x comma y ), where x (u comma v )space equals space 2 u plus space v squared, y (u comma v )space equals spa

barxatty [35]

$z=f(x(u,v),y(u,v)),\begin{cases}x(u,v)=2u+v^2\\y(u,v)=3u-v\end{cases}$

We're given that $f_x(6,1)=3$ and $f_y(6,1)=-1$ , and want to find $\frac{\partial z}{\partial v}(1,2)$ .

By the chain rule, we have

$\dfrac{\partial z}{\partial v}=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial v}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial v}$

and

$\dfrac{\partial x}{\partial v}=2v$

$\dfrac{\partial y}{\partial v}=-1$

Then

$\dfrac{\partial z}{\partial v}(1,2)=\dfrac{\partial z}{\partial x}(6,1)\dfrac{\partial x}{\partial v}(1,2)+\dfrac{\partial z}{\partial y}(6,1)\dfrac{\partial y}{\partial v}(1,2)$

(because the point $(x,y)=(6,1)$ corresponds to $(u,v)=(1,2)$ )

$\implies\dfrac{\partial z}{\partial v}(1,2)=3\cdot2\cdot2+(-1)\cdot(-1)=\boxed{13}$

4 0

2 years ago