All categories

3 years ago

Which relation is a function?

1 answer:

7 0

The definition of a function is that for each input, there is only one output. Thus, the answer to your question is A, as it is the only answer option that each input has only one output. In answer choice B, the input -1 has outputs of both -2 and -5. In answer choice C, the input 4 has outputs of both 3 and 0. Finally, in answer choice D, the input -4 has outputs of both -4 and 4. Therefore, answer A is the only function because it is the only option that follows the definition. Hope this helps! :)

You might be interested in

How do you add and subtract decimals as easily as you do whole numbers?

Gwar [14]

You can also multiply the decimals by a multiple of 10 to turn them I to whole numbers. Just make sure you multiply them both by the same multiple of 10 and divide them back before you write your answer!

8 0

3 years ago

5th grade math. correct answer will be marked brainliest

hodyreva [135]

Answer: C and for the bottom is 3*3/8=9/8=1 1/8

Step-by-step explanation: hoped this helped :)

8 0

3 years ago

Write the equation of the line in slope intercept form (y=mx+b) that passes through the point (-3,5) and has a slope of 2

MakcuM [25]

Answer:

no sorry

Step-by-step explanation:

for you question there needs to be a picture

4 0

2 years ago

Ayso has 18 8th grade boys teams this year but this is a 28% decrease from the prior year. How many 8th grade teams were there l

NikAS [45]

Answer:

23pupils

Step-by-step explanation:

Given parameters:

Number of 18th grade boys this year = 18

Percentage decrease = 28%

Unknown:

How many 8th grade teams were there last year = ?

Solution:

Since this is a percentage decrease, ;

Number of pupils this year = Number of pupils last year - (28% x 18 )

18 + (28% x 18 ) = Number of pupils last year

Number of pupils last year = 23pupils

4 0

2 years ago

What can this simplify to?

Mkey [24]

→ Chapter : Exponentiation ←
≡ We know that:
⇔ $\frac{a^{x}}{a^{y}}=a^{(x-y)}$
⇔ $\frac{1}{a^{-b}}=a^{b}$

≡ Solution:
⇒ $\frac{4^{2}g^{3}p^{-2}}{4^{5}g^{12}m^{-5}}$
⇒ $[4^{(2-5)}].[g^{(3-12)}].[p^{-2}].[m^{5}]$
⇒ $[4^{-3}].[g^{-9}].[p^{-2}].[m^{5}]$
∴ $\boxed{\frac{1}{64}g^{-9}p^{-2}m^{5}}$ or $\boxed{\frac{m^{5}}{64g^{9}p^{2}}}$

7 0

3 years ago