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3 years ago

Does the table represent a function?(Only 2 hours to answer!)

Mathematics

1 answer:

alisha [4.7K]3 years ago

6 0

Answer: (a) Yes. The table is a function.

<u>Step-by-step explanation:</u>

In order to be a function, there cannot be any duplicate x-values.

In the given table, each x-value is different so this IS a function.

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3. Find the mean and the median of this data set: 9,6,5, 3, 28, 6, 4, 7.<br>

BARSIC [14]

Answer:

Mean= 8.5

Median= 6

Step-by-step explanation:

3,4,5,6,6,7,9,28

Median= Middle (If two middle number, add and divide by 2)

Mean= (3+4+5+6+6+7+9+28) ÷8

7 0

3 years ago

Help asap !! will get branliest.

kompoz [17]

Answer:

0.8

Step-by-step explanation:

(2 x 10^-4)

= 0.0002

(4x10^3)

= 4000

0.0002 x 4000 = 0.8

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3 years ago

Find the equation of the line (in slope-intercept form) through the points (1,-4) and with slope 5/2.

Phoenix [80]

Answer:

Equation of a line is y = mx + c

Where m is the slope

c is the y intercept

Equation of the line using point

( 1 , - 4) and slope 5/2 is

$y + 4 = \frac{5}{2} (x - 1) \\ \\ y + 4 = \frac{5}{2} x - \frac{5}{2} \\ \\ y = \frac{5}{2} x - \frac{5}{2} - 4 \\ \\ \\ y = \frac{5}{2} x - \frac{13}{2}$

Hope this helps you

8 0

3 years ago

If Jorge walks 1 mile in 1 hour. In 2 hours at the same place, How many yards can Jorge walk now

julsineya [31]

Answer:

3520 yards

Step-by-step explanation:

1 mile in 1 hour.

In 2 hours he will cover 2×1= 2 miles

1 mile = 1760 yards

2 miles = 3520 yards

3 0

3 years ago

Read 2 more answers

Write a definite integral that represents the area of the region. (Do not evaluate the integral.) y1 = x2 + 2x + 3 y2 = 2x + 12F

Svet_ta [14]

Answer:

$A = \int\limits^3__-3}{9}-{x^{2}} \, dx = 36$

Step-by-step explanation:

The equations are:

$y = x^{2} + 2x + 3$

$y = 2x + 12$

The two graphs intersect when:

$x^{2} + 2x + 3 = 2x + 12$

$x^{2} = 0$

$x_{1} = 3\\x_{2} = -3$

To find the area under the curve for the first equation:

$A_{1} = \int\limits^3__-3}{x^{2} + 2x + 3} \, dx$

To find the area under the curve for the second equation:

$A_{2} = \int\limits^3__-3}{2x + 12} \, dx$

To find the total area:

$A = A_{2} -A_{1} = \int\limits^3__-3}{2x + 12} \, dx -\int\limits^3__-3}{x^{2} + 2x + 3} \, dx$

Simplifying the equation:

$A = \int\limits^3__-3}{2x + 12}-({x^{2} + 2x + 3}) \, dx = \int\limits^3__-3}{9}-{x^{2}} \, dx$

Note: The reason the area is equal to the area two minus area one is that the line, area 2, is above the region of interest (see image).

3 0

3 years ago