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mrs_skeptik [129]

2 years ago

13

If f(x) = x2, then f(4) = 42 = 16. Find: a. f(1) b. f(-3) c. f(t)

1 answer:

icang [17]2 years ago

4 0

Answer:

Step-by-step explanation:

f(x) = x²

f(4) = 4² = 16

f(1) = 1² = 1

f(-3) = -3² = 9

f(t) = t²

Always wise to us the ^ symbol to represent powers to avoid confusion

4^2 = 4² = 16

-3^2 = 9

t^2

if you are on a PC, the square symbol (²) can be typed by holding down the "alt" key while pressing and releasing "2","5","3" on the keypad, then release "alt". Unfortunately this does not work if you only have the row of numbers on the keyBoard to work with.

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Agora chess club has 16 members and gains a new memeber every month. The key club has 4 members and gains 4 new members every mo

svlad2 [7]

Answer:

For chess club:

y1 = 16 + 1*x where y1=total number of members after x months

For film club

y2 = 4 + 4*x where y2=total number of members after x months

y1 = y2 at

16 + x = 4 + 4x

3x = 12

x = 4

Therefore, after 4 months they the same number of members (equals 20 members, 16 + 1*4 or 4 + 4*4).

3 0

2 years ago

What is the value of the 5 in the number 7.522

joja [24]

Its 0.5. Which would be in fraction form 5/10 meaning its 5 tenths.

8 0

3 years ago

The mean age of a student book club is 14.2 years. A 27-year-old teacher is invited to join the club. How does the teacher’s age

MAXImum [283]

The new mean age will be greater than 14.2 years because the teacher is older than the previous mean.

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3 years ago

Read 2 more answers

A garden in the shape of a trapezoid has an area of 44.4 square

Leviafan [203]

Answer:

37 meters

Step-by-step explanation:

The equation for the area of a trapezoid is $A=\frac{a+b}{2} +h$ .

Here, $a = 4.3$ , $b=10.5$ , and $h$ is unknown.

$44.4=\frac{4.3+10.5}{2} +h$

$44.4=\frac{14.8}{2} +h$

$44.4=7.4 +h$

$37 = h$

The height of the trapezoid is the width of the garden, so the garden is 37 meters wide.

5 0

3 years ago

The Identity function i behaves just like the number 1. That is (F o i)=f and (i o g)= g

andrezito [222]

I'll denote the identity function by $\mathrm{Id}$ . Then for any functions $f$ with inverse $f^{-1}$ ,

$\begin{cases}f\circ\mathrm{Id}=f\\\mathrm{Id}\circ f=f\\f\circ f^{-1}=\mathrm{Id}\end{cases}$

One important fact is that composition is associative, meaning for functions $f,g,h$ , we have

$(f\circ g)\circ h=f\circ(g\circ h)$

So given

$h=f\circ g$

we can compose the functions on either side with $g^{-1}$ :

$h\circ g^{-1}=(f\circ g)\circ g^{-1}=f\circ(g\circ g^{-1})$

then apply the rules listed above:

$h\circ g^{-1}=f\circ\mathrm{Id}=f$

3 0

3 years ago