Please i really need help with this

Mathematics

2 answers:

solong [7]2 years ago

7 0

Answer:
V= 2x + 4
Step-by-step explanation:
First, we need to find out the original equation.
On this graph, we can see that the original y.
intercept is -1. We can also figure out the slope
by counting how many units up on the graph
the higher point is from the lower point, in this
case, 4, and how many units right it is, in this
case, 2, so our slope is 4/2, which can be
simplified to 2/1, or a slope of 2. Using the
slope and y-intercept, we can create the
equation for the original line. The equation will
be y = 2x - 1. To get our final answer, we simply
need to substitute 4 in the place of -1, and we
get our final equation, y = 2x + 4.

iris [78.8K]2 years ago

4 0

Answer:

y = 2x + 4

Step-by-step explanation:

First, we need to find out the original equation. On this graph, we can see that the original y-intercept is -1. We can also figure out the slope by counting how many units up on the graph the higher point is from the lower point, in this case, 4, and how many units right it is, in this case, 2, so our slope is 4/2, which can be simplified to 2/1, or a slope of 2. Using the slope and y-intercept, we can create the equation for the original line. The equation will be y = 2x - 1. To get our final answer, we simply need to substitute 4 in the place of -1, and we get our final equation, y = 2x + 4.

You might be interested in

What is the measurement of the longest line segment in a right rectangular prism that is 26 inches long, 2 inches wide, and 2 in

EastWind [94]

Answer:

$6\sqrt{19} \approx 26.153$ inches.

Step-by-step explanation:

The longest line segment in a right rectangular prism is the diagonal that connects two opposite vertices. On the first diagram attached, the green line segment connecting A and G is one such diagonals. The goal is to find the length of segment $\mathsf{AG}$ .

In this diagram (not to scale,) $\mathsf{AB} = 26$ (length of prism,) $\mathsf{AC} = 2$ (width of prism,) $\mathsf{AE} = 2$ (height of prism.)

Pythagorean Theorem can help find the length of $\mathsf{AG}$ , one of the longest line segments in this prism. However, note that this theorem is intended for right triangles in 2D, not the diagonal in a 3D prism. The workaround is to simply apply this theorem on two different right triangles.

Start by finding the length of line segment $\mathsf{AD}$ . That's the black dotted line in the diagram. In right triangle $\triangle\mathsf{ABD}$ (second diagram,)

Segment $\mathsf{AD}$ is the hypotenuse.
One of the legs of $\triangle\mathsf{ABD}$ is $\mathsf{AB}$ . The length of $\mathsf{AB}$ is $26$ , same as the length of this prism.
Segment $\mathsf{BD}$ is the other leg of this triangle. The length of $\mathsf{BD}$ is $2$ , same as the width of this prism.

Apply the Pythagorean Theorem to right triangle $\triangle\mathsf{ABD}$ to find the length of $\mathsf{AB}$ , the hypotenuse of this triangle:

$\mathsf{AD} = \sqrt{\mathsf{AB}^2 + \mathsf{BD}^2} = \sqrt{26^2 + 2^2}$ .

Consider right triangle $\triangle \mathsf{ADG}$ (third diagram.) In this triangle,

Segment $\mathsf{AG}$ is the hypotenuse, while
$\mathsf{AD}$ and $\mathsf{DG}$ are the two legs.

$\mathsf{AD} = \sqrt{26^2 + 2^2}$ . The length of segment $\mathsf{DG}$ is the same as the height of the rectangular prism, $2$ (inches.) Apply the Pythagorean Theorem to right triangle $\triangle \mathsf{ADG}$ to find the length of the hypotenuse $\mathsf{AG}$ :

$\begin{aligned}\mathsf{AG} &= \sqrt{\mathsf{AD}^2 + \mathsf{GD}^2} \\ &= \sqrt{\left(\sqrt{26^2 + 2^2}\right)^2 + 2^2}\\ &= \sqrt{\left(26^2 + 2^2\right) + 2^2} \\&= 6\sqrt{19} \\&\approx 26.153\end{aligned}$ .