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2 years ago

Solve this system of equations by using the elimination method. 2x-3y=25 5x+3y=10

1 answer:

7 0

Elimination method is my favorite way to solve 2 equations like this written in standard form! First I'm going to rewrite the equations and mark numbers to make this explanation more clear.

2x-3y=251️⃣

5x+3y=102️⃣

Looking at these 2 equations we can see that there is a +3 and a -3. This could be eliminated by adding the 2 equations together! *Remember what is on the left side adds with the left side, and what's on the right side adds with the right side. Therefore we can do:

1️⃣+2️⃣

2x-3y+5x+3y=25+10

7x=35

x=5

Then we can substitute the value of x back into either of the 2 equations.

2*5-3y=25

10-3y=25

3y=-15

y=-5

Then write the 2 answers together.

{x=5

{

{y=-5

Now we can check our answers.

2*5-3*-5=10- -15= 10+15=25 Correct

5*5+3*-5= 25+ -15= 10 Correct

From this we can see that x=5 y= - 5 are the correct answers!

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Step-by-step explanation:

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3 years ago

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2 years ago

Function or not a function?

Mazyrski [523]

<h3><u>Explanation</u></h3>

How to see if a shown graph is function or not?

If we want to check that a graph is function or not, we have a way to check by doing these steps.

Draw a vertical line, make sure that a line has to pass through or intercept a graph.
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If a line intercepts a graph only one point, a graph is indeed a function. Otherwise, not a function but a relation instead. That includes if a line intercepts more than a point which doesn't make a graph a function.

From the graph, if we follow these steps, we will see that a line will only pass or intercept the graph only one point. Hence, the graph is indeed a function. The following graph that is shown is called "Parabola" for a < 0.

<h3><u>Answer</u></h3>

The graph is a function.

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2 years ago

Describe the location of the point having the following coordinates. negative abscissa, zero ordinate between Quadrant II and Qu

Marianna [84]

Answer:

The location of the point is between Quadrant II and Quadrant III

Step-by-step explanation:

we know that

The abscissa refers to the x-axis and ordinate refers to the y-axis

in this problem we have

the coordinates of the point are $(-x,0)$

see the attached figure to better understand the problem

The location of the point is between Quadrant II and Quadrant III

6 0

2 years ago

A second particle, Q, also moves along the x-axis so that its velocity for 0 £ £t 4 is given by Q t cos 0.063 ( ) t 2 v t( ) = 4

Vladimir79 [104]

Answer:

The time interval when $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

The distance is 106.109 m

Step-by-step explanation:

The velocity of the second particle Q moving along the x-axis is :

$V_{Q}(t)=45\sqrt{t} cos(0.063 \ t^2)$

So ; the objective here is to find the time interval and the distance traveled by particle Q during the time interval.

We are also to that :

$V_Q(t) \geq 60$ between $0 \leq t \leq 4$

The schematic free body graphical representation of the above illustration was attached in the file below and the point when $V_Q(t) \geq 60$ is at 4 is obtained in the parabolic curve.

So, $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

Taking the integral of the time interval in order to determine the distance; we have:

distance = $\int\limits^{3.519}_{1.866} {V_Q(t)} \, dt$

= $\int\limits^{3.519}_{1.866} {45\sqrt{t} cos(0.063 \ t^2)} \, dt$

= By using the Scientific calculator notation;

distance = 106.109 m

4 0

3 years ago