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3 years ago

What are the intercepts of f(x)=x^2

Mathematics

1 answer:

dusya [7]3 years ago

7 0

C. It is a basic parabola

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How many points need to be removed from this graph so it will be a function?

finlep [7]

Answer:

3 points

Step-by-step explanation:

in a function, an x-value must have only one y-value correlated with it.

So, you would need to remove either (-3, 2) or (-3, 3)

And, you would need to remove either (2, -2) or (2, -4)

<em>And, </em>you would need to remove either (1, 4) or (1, -4)

So--you need to remove 3 points in total for this to be a function.

[a y-value can have multiple x-values. the two points at -4 are allowed, and do not need to be removed for both of them to be there]

8 0

1 year ago

Simplify the expression 17y-2x+3(3x-y)+2x

Marat540 [252]

Answer:

9x+14y

Step-by-step explanation:

First Multiply the number outside of the brackets with the numbers inside the brackets. Ex:3 x 3x and 3 x -y

17y-2x+3(3x-y)+2x

That should get you to this.

17y-2x+9x-3y+2x

Next combine like terms.

14y+9x

Hope this helps!

5 0

3 years ago

If the pet store had 10 more birds the number of dogs would be double the number of birds what numbers should be on the scale

frosja888 [35]

If there were 10 dogs then there would be 20 birds

4 0

3 years ago

Read 2 more answers

What is 3/5 equal to ?

Crazy boy [7]

The correct answer to your question is 0.6 !

4 0

3 years ago

Read 2 more answers

A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee

katrin [286]

Answer: A) 1260

Step-by-step explanation:

We know that the number of combinations of n things taking r at a time is given by :-

$^nC_r=\dfrac{n!}{(n-r)!r!}$

Given : Total multiple-choice questions = 9

Total open-ended problems=6

If an examine must answer 6 of the multiple-choice questions and 4 of the open-ended problems ,

No. of ways to answer 6 multiple-choice questions

= $^9C_6=\dfrac{9!}{6!(9-6)!}=\dfrac{9\times8\times7\times6!}{6!3!}=84$

No. of ways to answer 4 open-ended problems

= $^6C_4=\dfrac{6!}{4!(6-4)!}=\dfrac{6\times5\times4!}{4!2!}=15$

Then by using the Fundamental principal of counting the number of ways can the questions and problems be chosen = No. of ways to answer 6 multiple-choice questions x No. of ways to answer 4 open-ended problems

= $84\times15=1260$

Hence, the correct answer is option A) 1260

5 0

3 years ago