Set of ordered pairs: Is it a function? (yes no ) Why or why not? (0,3),(-1,4),(2,4),(0,8),(5,7)

Mathematics

1 answer:

kari74 [83]2 years ago

8 0

Answer:

Step-by-step explanation:

A function cannot have two different points with the same x-coordinate. For instance in this one, (0,3) & (0,8) share the same value of x

You might be interested in

Mark shoveled 1/8 ton of gravel in 1/4 hour. What was his average rate of shoveling in tons per hour?

mel-nik [20]

Answer:

1/2 tons of gravel per hour

Step-by-step explanation:

1/8=1/4

2/8=2/4

3/8=3/4

4/8 or 1/2 = 4/4 or 1

4 0

3 years ago

Write the next three terms in the pattern. What is each pattern rule?<br><br> +6, -12, +18, -24, ...

natali 33 [55]

Answer: 30, -36, 42

Step-by-step explanation:

So basically the pattern is just the next number that is a multiple of six, but every other number is negative. 6 is 6 times 1, -12 is 6 times -2, 18 is 6 times 3, -24 is 6 times -4, 30 is 6 times 5, -36 is 6 times -6, and 42 is 6 times 7

7 0

2 years ago

If a trapezoid with 9.00-cm and 12.00-cm bases has an area of 68.25 cm2, what is its height to the nearest hundredth?

Lapatulllka [165]

I hope this helps you

68,25=(9+12).h/2

2.68,25=21.h

136,5/21=h

h=6,5

4 0

3 years ago

What is the simplified form of the equation fraction 4 over 5 n minus fraction 1 over 5 equals fraction 2 over 5 n

liubo4ka [24]

4/5n – 1/5 = 2/5n
Add 1/5 to both sides
4/5n = 2/5n + 1/5
Subtract 2/5n from both sides
2/5n = 1/5
Divide both sides by 2/5

n = 1/2

8 0

3 years ago

Express p in terms of q if

Irina-Kira [14]

$p=x^2+\dfrac{1}{x^2}\\\\p=\dfrac{x^4}{x^2}+\dfrac{1}{x^2}\\\\p=\dfrac{x^4+1}{x^2}\qquad(*)$

$q=x+\dfrac{1}{x}\\\\q=\dfrac{x^2}{x}+\dfrac{1}{x}\\\\q=\dfrac{x^2+1}{x}\qquad\text{square both sides}\\\\q^2=\left(\dfrac{x^2+1}{x}\right)^2\\\\q^2=\dfrac{(x^2+1)^2}{x^2}\qquad\text{use}\ \ (a+b)^2=a^2+2ab+b^2\\\\q^2=\dfrac{(x^2)^2+2(x^2)(1)+1^2}{x^2}\\\\q^2=\dfrac{x^4+2x^2+1}{x^2}\\\\q^2=\dfrac{x^4+1}{x^2}+\dfrac{2x^2}{x^2}\\\\q^2=\dfrac{x^4+1}{x^2}+2\qquad\text{subtract 2 from both sides}\\\\q^2-2=\dfrac{x^4+1}{x^2}\\\\\text{From (*) we have}\\\\\boxed{p=q^2-2}$

8 0

2 years ago