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Novay_Z [31]
3 years ago
11

At genesee high school, the sophomore class has 60 more students than the freshman class

Mathematics
1 answer:
Igoryamba3 years ago
3 0

coooooooooooooooooooooooooooooool

You might be interested in
42 add 1 than multiply by 4
AleksandrR [38]

Answer:

172

Step-by-step explanation:

42+1= 43

43 x 4= 172

5 0
3 years ago
Read 2 more answers
The area of a square is defined by, A(x) = x 2 - 6x + 9. What is the length of a side of the square?
Dmitriy789 [7]
A9x)  = x^2 - 6x + 9  = (x - 3)^2

length of a side = x - 3 
8 0
3 years ago
Taletha is preparing for an exam in Algebra II. She knows she will be expected to determine if two functions are inverses of eac
seropon [69]

Answer:

f ○ g(x) = x and g ○ f(x) = x  is the correct composition

Step-by-step explanation:

INVERSE FUNCTIONS: Two functions are said to be inverse of each other if for every y = f(x), there exists a function g(x),

such that x = f^{-1} (y) =  g(y)

Now, here if we need to show tha two functions f(x) and g(x) are inverse functions of each other , then show that for every x:

1. f  o g(x)   =   f(g(x))= x

2. g o f(x)  =  g(f(x))   = x

If any two functions satisfy these two conditions, then they are inverse of each other.

5 0
4 years ago
Let f be the function given by f(x)= (x-1)(x^2-4)/x^2-a. For what positive values of a is f continuous for all real numbers x?
9966 [12]

Answer:

a =1 and a=4.

Step-by-step explanation:

The function is

f(x)=\frac{(x-1)(x^2-4)}{x^2-a}

If we want f(x) to be continuous the denominator needs to be different to 0, otherwise f(x) will be indeterminate.

Now, for a a positive real we have that x=\sqrt{a} will annulate the denominator, i.e

(\sqrt{a})^2-a = a-a = 0. But, if a = 1 we have:

f(x)=\frac{(x-1)(x^2-4)}{x^2-1} = \frac{(x-1)(x^2-4)}{(x-1)(x+1)}=\frac{(x^2-4)}{x+1}

so, the value x=\sqrt{a} = \sqrt{1} = 1 won't annulate the denominator.

Now, for a = 4 we have:

f(x)=\frac{(x-1)(x^2-4)}{x^2-4} = x-1

so, the value x=\sqrt{a} = \sqrt{4} = 2 won't annulate the denominator.

In conclusion, for a=1 or a=1, the function will be continuos for all real numbers, since the denominator will never be 0.

4 0
3 years ago
What are roots of the equation -90=x (x-18)
daser333 [38]

Answer:

-90=x(x-18)

-90=x^{2}-18x

0=x^{2}-18x+90

x=9+3i   or   9-3i

Step-by-step explanation:

you need to use quadratic formula and you will find square root a negative ( -9 ), which is an imaginary no.

square root (-9) = 3i

4 0
3 years ago
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