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horsena [70]
3 years ago
9

Let f(x) = 2x2 and g(x) = x2-1. Then f(g(2)) =

Mathematics
1 answer:
finlep [7]3 years ago
5 0

Answer:

18

Step-by-step explanation:

Evaluate g(2), then substitute the value obtained into f(x)

g(2) = 2² - 1 = 4 - 1 = 3 , then

f(3) = 2(3)² = 2(9) = 18

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Lin parachutes from a plane at a height of 22,000ft above sea level. Several minutes later she notes from an instrument on her w
Brums [2.3K]
22000 - 10000 = 12000ft

Hope this helps! :)


4 0
3 years ago
Read 2 more answers
At benitos school 5/8 of the students like math. If there are 208 students, how many like math class
Rom4ik [11]

Answer:

`130 students like math

Step-by-step explanation:

4 0
3 years ago
Let x = a bi and y = c di and z = f gi which statements are true? check all of the boxes that apply. x y = y x (x × y) × z = x ×
Alex73 [517]

A complex number is a part of a number system that includes a real unit and an imaginary unit. All the statements are true except C and E.

<h3>What is a Complex Number?</h3>

A complex number is a part of a number system that includes the real numbers as well as a particular element labelled i, sometimes known as the imaginary unit, and which obeys the equation i² = 1.

As it is given that x=(a+bi), y=(c+di), and z=(f+gi), therefore, we need to plug in the values to check which all are true.

A.) x+y=y+x

x+y=y+x\\a+bi+c+di=c+di+a+bi

As the two sides of the equation are equal, the equation is true.

B.) (x × y) × z = x × (y × z)

(x \times y) \times  z = x \times  (y \times  z)\\\\\ [ (a+bi)  \times (c+di) ]\times (f+gi) = (a+bi)  \times [(c+di) \times (f+gi)]\\\\(ac+adi+bci-bd) \times (f+gi)=  (a+bi)  \times (cf+cgi+fdi-dg)

(ac+adi+bci-bd) \times (f+gi)=  (a+bi)  \times (cf+cgi+fdi-dg)\\\\\\acf+adfi+bcfi-bdf+acgi-adg-bcg-bdgi = acf+acgi+afdi-adg + bcfi-bcg-bfd-bdgi

As the two sides of the equation are equal, the equation is true.

C.) x−y=y−x

x-y=y-x\\\\(a+bi)-(c+di)=(c+di)-(a+bi)\\\\a+bi-c-di=c+di-a-bi

As the two sides of the equation are not equal, the equation is not true.

D.) (x+y)+z=x+(y+z)

(x+y)+z=x+(y+z)\\\\(a+bi+c+di)+f+gi = a+bi+(c+di+f+gi)\\\\a+bi+c+di+f+gi = a+bi+c+di+f+gi

As the two sides of the equation are equal, the equation is true.

E.) (x−y)−z=x−(y−z)

(x-y)-z=x-(y-z)\\\\(a+bi-c-di)-f-gi = a+bi-(c+di-f-gi)\\\\a+bi-c-di-f-gi =a+bi-c-di+f+gi)

As the two sides of the equation are not equal, the equation is not true.

Hence, all the statements are true except C and E.

Learn more about Complex Number:

brainly.com/question/12464608

6 0
2 years ago
What is the product? 6(x*2-1)(6x-1/6(x+1))
Yanka [14]
6(x^2-1)(6x-1/6(x+1))

On factoring (x^2-1)=(x+1)(x-1)

= 6(x-1)(x+1)(6x-1/6(x+1))
Cancelling (x+1) from numerator and denominator we get,

= 6(x-1)(6x-1)/6

Cancelling 6 we get,
= (x-1)(6x-1)
4 0
3 years ago
In how many ways can 13 people be divided into four groups with 3, 6, 2 and 2 people respectively?
OleMash [197]

Answer:

Total combination = 180,180

Step-by-step explanation:

It is a combination based problem.

So,

Total people = 13

Group = 3,6,2,2

So,

For 6 people

13C6= 1,716

For 3 people

7C3 = 35

For 2 group of 2 people

(4C2)/2 = 3

Total combination = 1,716 × 35 × 3

Total combination = 180,180

8 0
3 years ago
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