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Anettt [7]
2 years ago
15

Richard is frustrated. He knows the line to the right should have a 3 and a 4 in the equation but he can't remember if it is y =

4x + 3 or y = 3x + 4. a. Which equation is correct? How do you know?
Mathematics
1 answer:
Alisiya [41]2 years ago
7 0

Answer:

The correct equation is "y=4x+3"

Step-by-step explanation:

Given equation:

y= 4x+3 \ \ or \ \ y=3x+4

As per the given line formula:

\to y=mx+c

The slope(m) of the line is \frac{4}{1} = 4  and y-intercept at 3, that is the point (c). So, the correct equation of the line:   y = 4x + 3

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aksik [14]

1+1 = 84, explanation is quite simple. i should get brainliiest

4 0
3 years ago
!!! CAN SOMEONE HELP !!! (PLEASE ILL GIVE BRAINLIST)
vladimir1956 [14]

Answer:

120 minutes

Step-by-step explanation:

the total bill was $58 so we have to subtract the $40 monthly bill part to find how much extra money she paid because her call was over 200 minutes

58-40=18

Then divide the $18 by $0.15 since the $18 is the total amount of each time she was charged $0.15 for going 1 minute over 200 minutes

18÷0.15 = 120

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6 0
3 years ago
Read 2 more answers
DIscrete Math
Daniel [21]

Answer:

Step-by-step explanation:

As the statement is ‘‘if and only if’’ we need to prove two implications

  1. f : X \rightarrow Y is surjective implies there exists a function h : Y \rightarrow X such that  f\circ h = 1_Y.
  2. If there exists a function h : Y \rightarrow X such that  f\circ h = 1_Y, then f : X \rightarrow Y is surjective

Let us start by the first implication.

Our hypothesis is that the function f : X \rightarrow Y is surjective. From this we know that for every y\in Y there exist, at least, one x\in X such that y=f(x).

Now, define the sets X_y = \{x\in X: y=f(x)\}. Notice that the set X_y is the pre-image of the element y. Also, from the fact that f is a function we deduce that X_{y_1}\cap X_{y_2}=\emptyset, and because  f the sets X_y are no empty.

From each set X_y  choose only one element x_y, and notice that f(x_y)=y.

So, we can define the function h:Y\rightarrow X as h(y)=x_y. It is no difficult to conclude that f\circ h(y) = f(x_y)=y. With this we have that f\circ h=1_Y, and the prove is complete.

Now, let us prove the second implication.

We have that there exists a function  h:Y\rightarrow X  such that f\circ h=1_Y.

Take an element y\in Y, then f\circ h(y)=y. Now, write x=h(y) and notice that x\in X. Also, with this we have that f(x)=y.

So, for every element y\in Y we have found that an element x\in X (recall that x=h(y)) such that y=f(x), which is equivalent to the fact that f is surjective. Therefore, the prove is complete.

3 0
3 years ago
If the answer to the square root is 5, then what is the number that gets you 5
tensa zangetsu [6.8K]
\sqrt x=5\\
x=25

4 0
3 years ago
At which points are the tangents drawn to the ellipse x 2 + y 2 = [ a ] x + [ a ] y parallel to
In-s [12.5K]

The given equation of the ellipse is x^2 + y^2 = 2 x + 2 y

At tangent line, the point is horizontal with the x-axis therefore slope = dy / dx = 0

<span>So we have to take the 1st derivative of the equation then equate dy / dx to zero.</span>

x^2 + y^2 = 2 x + 2 y

x^2 – 2 x = 2 y – y^2

(2x – 2) dx = (2 – 2y) dy

(2x – 2) / (2 – 2y) = 0

2x – 2 = 0

x = 1

 

To find for y, we go back to the original equation then substitute the value of x.

x^2 + y^2 = 2 x + 2 y

1^2 + y^2 = 2 * 1 + 2 y

y^2 – 2y + 1 – 2 = 0

y^2 – 2y – 1 = 0

Finding the roots using the quadratic formula:

y = [-(- 2) ± sqrt ( (-2)^2 – 4*1*-1)] / 2*1

y = 1 ± 2.828

y = -1.828 , 3.828

 

<span>Therefore the tangents are parallel to the x-axis at points (1, -1.828) and (1, 3.828).</span>

3 0
3 years ago
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