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We find the value of f(x) that results in 3(f(x)=0). 

For 3-x, x=3. For -x^3+2x^2+3x, x=0, 3, -1. None of these satisfies x>3, so f(0) cannot go into the second piece of the function. It can only go in the first, so f^-1(0) is 3.

Then we find the value of f(x) that results in 6. 

For 3-x, x=-3. For -x^3+2x^2+3x, x=2, \(\sqrt{3}\ \text{or}-\sqrt{3}\). None of these satisfies x>3, so again, f(6) cannot go into the second piece of the function. It can only go in the first, so f^-1(6) is -3.

Adding these together, 3+(-3)=0.

Help you with what?

Betsy must have one quarter and 4 nickels ( for Andrew to have a multiple of her quarters)

Andrew has 2 quarters and 2 nickels = 60 cents

X = Apple price

she paid 100% of x.  Plus 3% of x.     Or. 1.03x

1.03x = 19.16.   Divvied both sides by 1.03

x= 19.16/1.03=......

If R is the radius of the circle: White area is (2R*2R)/2, which is the same as 2R^2.

For the orange area, we can do 2R*2R-3.14R^2, which would be the same as 0.86R^2. 

The blue area, we do 3.14*R^2-2R^2, which is the same as 1.13R^2.

So, we can see that the blue is bigger.

Hope this helped!

The sum of the ages of Anna, Bernard and Carl is 32 years. Bernard is 4 years older than Anna, and Carl is 6 years older than Bernard. What is Anna age?

Hello Guest!
 

a+b+c=32

a+4=b           

b+6=c  

a+b+c=32

(b-4)+b+(b+6)=32

3b=32+4-6

3b=30

b=10

c=16

a+10+16=32

a=32-10-16

a=6

Anna is 6, Bernard is 10, Carl is 16

  !

a=1;b=floor(a/6);c=floor(a/7) ; if(b==c, goto4, goto5);printa,b, c; a++;if(a<1000, goto1, 0)

n   n/6  n/7

1     0     0
2    0     0
3    0     0
4    0     0
5    0     0
7     1     1
8     1    1
9     1    1
10   1    1
11    1    1
14   2    2
15   2    2
16   2    2
17   2    2
21   3    3
22  3    3
23  3    3
28  4    4
29  4    4
35  5    5
 

thank you so much!!!!

but what if i dont have an account

1 + i  is  not real, so  f(1 + i)  =  (1 + i)²  =  1² + 2i + i²  =  1 + 2i - 1  =  2i

2i  is  not real, so  f( f(1 + i) )  =  f(2i)  =  4i²  =  -4

-4  is  real, so  f( f( f(1 + i) ) )  =  -(-4)²  =  -16

-16  is  real, so  f( f( f( f(1 + i) ) ) )  =  -(-16)²  =  -256  

This could be your AP:

- 1/2,  2,  3,  4,  5,  6,  7,  8,  9,  10

Surprise! Your calculation is what Wolfram/Alpha also gets. Go to this link and enter your formulas in there and see what you get. I don't know much about chemistry.

https://www.wolframalpha.com/input/?i=Calculate+pH++pKa+%3D+10.25%2C++25.2+mmol%2FL%2C++32.0+umol%2FL+

This has been posted and answered recently....search for it...

Explanation:

There are 56 ways to choose 2 girls out of 8 and there are 7 ways to choose 1 boy out of 7. 

Because we use the word "and", we multiply 56 by 7, and we get 392.

But then, there's the possibility that we choose 1 girl and 2 boys.

There are 8 ways to choose 1 girl out of 8 and there are 42 ways to choose 2 boys out of 7.

Again, we multiply due to the word "and", so we get 336

We can add 392 and 336 because we choose 2 girls and 1 boy OR 1 girl and 2 boys.

Finally, we get an answer of 728.

Solution #2:

You can choose 1 girl and 1 boy first.

There are 8 ways to choose 1 girl and 7 ways to choose 1 boy. 

Now there are 13 students left and one more student needed to help.

There are 13 ways to choose 1 more student. Gender doesn't matter now as we already have 1 girl and 1 boy.

Now we multiply as we need to choose 1 girl, 1 boy, AND another student.

Once again, we get an answer of 728.

fix your latex please

There are numbers that are missing.

1. 270 counterclockwise: (x,y)-->(y,-x)

So (3,-4)--> a.(-3,-4)

2. R is 4 units below the x axis. When reflected, it is 4 untis above the x axis. So R' would be a. (3, 4)

3. Translating 13 units up is essentially the same as adding 13 to the y point so -4+13=9. So R' would be d. (3,9).

4. A scale factor of 1/5 results in 1/5 of all the points.

L(0,10)-->L'(0,2)

M(-5, -5)-->M'(-1,-1)

J(0,-10)-->J'(0,-2)

K(5,-5)-->K'(1,-1)

So the answer is b.

For f(2)=5, 2 is greater than 0 so it's substituted into the equation ax+3 to get 2a+3=5.

For f(0)=5, 0 is directly equal to 0 so it's substituted into the equation ab to get ab=5.

For f(-2)=-10, -10 is less than 0 so it's substituted into the equation bx+c to get -2b+c=-10.

We have a system of equations now:

2a+3=5

ab=5

-2b+c=-10

From the first equation, 2a+3=5, a=1.

We substitute the value of a into ab=5, and get b=5.

Finally, from the last equation, -2b+c=-10, c=0.

So a+b+c=6.

Then Guesswork...

5+9= 14?

That's the best I can do.

I don't think it's correct but based on the question that's the only thing I can do.

(e_1,e_2) works out to (4,6).

(e_1,e_2) works out to (5,6).

We are given the expressions of each angle and we know the angles in a quadrilateral add up to 360.

So we form an equation:

q+5q+15+3q-25+4q-20=360

13q-30=360

13q=390

q=30

The largest angle is 5q+15, which is 5(30)+15=\(165^\circ\).

I think the image you linked is local to your computer, so we can't see it...