What is the solution to the system of linear equations ?

For which value of X is the row in the table of values incorrect? The function is the quadratic function Y=3/4x^2

svet-max [94.6K]

Answer:

The row with x = 2 and y = 4 is incorrect. The y value should be y = 3 when x = 2.

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Explanation:

Fractions can be a bit of a pain, so I'm going to be converting to decimal form to help with the computation; however, I will convert the y value results to mix number form so that it matches what the table shows.

Keep in mind that 3/4 = 0.75 when you use a calculator or long division. So a number like 2.75 means 2 & 3/4 (two and three quarters)

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Let's check the first row. Plug in x = 0 and it leads to...

y = (3/4)*x^2

y = (3/4)*0^2

y = (3/4)*0

y = 0.75*0

y = 0

This shows us that row 1 is correct. When x = 0, y = 0 leading to (x,y) = (0,0).

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Repeat the same basic outline of steps, but now use x = 2

y = (3/4)*x^2

y = (3/4)*2^2

y = (3/4)*4

y = 0.75*4

y = 3

So instead of 4, the proper y value is y = 3 when x = 2. This means row 2 is the answer (the row that has the incorrect values).

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For the sake of completeness, let's see if the third row is correct or not. Plug in x = 3

y = (3/4)*x^2

y = (3/4)*3^2

y = (3/4)*9

y = 0.75*9

y = 6.75

y = 6 & 3/4

We see that x = 3 points to y = 6 & 3/4, so we have the correct values in this row. We can cross this off the list of choices.

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Finally, let's check x = 5

y = (3/4)*x^2

y = (3/4)*5^2

y = (3/4)*25

y = 0.75*25

y = 18.75

y = 18 & 3/4

We get the proper y value. So we can also cross this off the list.

5 0

2 years ago

9. The ratio of Joe's weight to Colin's is 2:3. Colin's weight is 36kg. If Joe gains

STatiana [176]

2:3
?:36

36/3=13
2x13=26

26:36

26+3=29

29:36

6 0

2 years ago

Which of the following is true?

Oksana_A [137]

B is true

Step-by-step explanation:

because no number with them to lines can be a negative

4 0

2 years ago

Read 2 more answers

The equation y=9. 5x 0. 4 represents the line of best fit for the distance (in feet) that Marcella traveled over time (in second

Bas_tet [7]

The average speed of Marcella in feet per seconds is 9.5 feet per second.

<h3>How are time taken for travel, distance traveled, and the average speed are related?</h3>

We have this below shown relation between them

$\text{Average speed} = \dfrac{\text{Distance traveled}}{\text{Total time taken to travel that distance}}\\\\or\\\\\text{Total time taken to travel that distance}= \dfrac{\text{Distance traveled}}{\text{Average speed} }$

For the given case, we have:

The equation $y=9.5x + 0.4$ represents the line of best fit for the distance (in feet) that Marcella traveled over time (in seconds)

The line of best fit can be taken as approximately telling about the data points ( y = distance traveled in feet, and x = time spent )

Let the initial time be 'x' with initial distance traveled 'y', then the time after one second will be (x+1), and the distance traveled be $y_1$

The value of y and $y_1$ are calculated as:

$y = 9.5x + 0.4\\y_1 = 9.5(x+1) + 0.4 = 9.5x + 9.9$

Thus, the average speed is calculated at time 'x' as:

$\text{Average speed} = \dfrac{\text{Distance traveled}}{\text{Total time taken to travel that distance}}\\\\or\\\\\text{Average speed} = \dfrac{y_1 - y}{x + 1 - x} = \dfrac{9.5x + 9.9 - 9.5x - 0.4}{1} = 9.5 \: \rm feet/sec$