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Stels [109]
3 years ago
5

Which fraction has a terminating decimal as its decimal expansion?

Mathematics
2 answers:
Irina18 [472]3 years ago
8 0
Yes i agree i think so
ArbitrLikvidat [17]3 years ago
3 0

Answer:

fraction which has a terminating decimal as its decimal expansion is:

1/5

Step-by-step explanation:

We are given 4 fractions:

1/3 , 1/5 , 1/7 and 1/9

We have to find which fraction has terminating decimal.

1/3 = 0.33333...

1/5 = 0.2

1/7 = 0.142857142857...

1/9 =0.11111.....

Hence, fraction which has a terminating decimal as its decimal expansion is:

1/5

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no

Step-by-step explanation:

this is because when dealing with the power, it involves multiplication

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Given that f.x 3x-2 over x+1 g[x] x +5 evaluate f[-4] and gf [-2]
Jobisdone [24]

The value of f[ -4 ] and g°f[-2] are \frac{14}{3} and 13 respectively.

<h3>What is the value of f[-4] and g°f[-2]?</h3>

Given the function;

  • f(x) = \frac{3x-2}{x+1}
  • g(x)=x+5
  • f[ -4 ] = ?
  • g°f[ -2 ] = ?

For f[ -4 ], we substitute -4 for every variable x in the function.

f(x) = \frac{3x-2}{x+1}\\\\f(-4) = \frac{3(-4)-2}{(-4)+1}\\\\f(-4) = \frac{-12-2}{-4+1}\\\\f(-4) = \frac{-14}{-3}\\\\f(-4) = \frac{14}{3}

For g°f[-2]

g°f[-2] is expressed as g(f(-2))

g(\frac{3x-2}{x+1}) =  (\frac{3x-2}{x+1}) + 5\\\\g(\frac{3x-2}{x+1}) =  \frac{3x-2}{x+1} + \frac{5(x+1)}{x+1}\\\\g(\frac{3x-2}{x+1}) =  \frac{3x-2+5(x+1)}{x+1}\\\\g(\frac{3x-2}{x+1}) =  \frac{8x+3}{x+1}\\\\We\ substitute \ in \ [-2] \\\\g(\frac{3x-2}{x+1}) =  \frac{8(-2)+3}{(-2)+1}\\\\g(\frac{3x-2}{x+1}) =  \frac{-16+3}{-2+1}\\\\g(\frac{3x-2}{x+1}) =  \frac{-13}{-1}\\\\g(\frac{3x-2}{x+1}) =  13

Therefore, the value of f[ -4 ] and g°f[-2] are \frac{14}{3} and 13 respectively.

Learn more about composite functions here: brainly.com/question/20379727

#SPJ1

6 0
2 years ago
Given that a:b:c= 20 : 35 : 15, a) simplify a : b:c, b) find a:b,
Furkat [3]

Answer:

a) 4:7:3

b) 4:7 / 20:35

c) 3:7 / 15:35

4 0
2 years ago
If(x) = x + 2 and h(x) = x-1, what is f • h](-3)?
deff fn [24]

Answer/Step-by-step explanation:

Composition functions are functions that combine to make a new function. We use the notation ◦ to denote a composition.

f ◦ g is the composition function that has f composed with g. Be aware though, f ◦ g is not

the same as g ◦ f. (This means that composition is not commutative).

f ◦ g ◦ h is the composition that composes f with g with h.

Since when we combine functions in composition to make a new function, sometimes we

define a function to be the composition of two smaller function. For instance,

h = f ◦ g (1)

h is the function that is made from f composed with g.

For regular functions such as, say:

f(x) = 3x

2 + 2x + 1 (2)

What do we end up doing with this function? All we do is plug in various values of x into

the function because that’s what the function accepts as inputs. So we would have different

outputs for each input:

f(−2) = 3(−2)2 + 2(−2) + 1 = 12 − 4 + 1 = 9 (3)

f(0) = 3(0)2 + 2(0) + 1 = 1 (4)

f(2) = 3(2)2 + 2(2) + 1 = 12 + 4 + 1 = 17 (5)

When composing functions we do the same thing but instead of plugging in numbers we are

plugging in whole functions. For example let’s look at the following problems below:

Examples

• Find (f ◦ g)(x) for f and g below.

f(x) = 3x + 4 (6)

g(x) = x

2 +

1

x

(7)

When composing functions we always read from right to left. So, first, we will plug x

into g (which is already done) and then g into f. What this means, is that wherever we

see an x in f we will plug in g. That is, g acts as our new variable and we have f(g(x)).

g(x) = x

2 +

1

x

(8)

f(x) = 3x + 4 (9)

f( ) = 3( ) + 4 (10)

f(g(x)) = 3(g(x)) + 4 (11)

f(x

2 +

1

x

) = 3(x

2 +

1

x

) + 4 (12)

f(x

2 +

1

x

) = 3x

2 +

3

x

+ 4 (13)

Thus, (f ◦ g)(x) = f(g(x)) = 3x

2 +

3

x + 4.

Let’s try one more composition but this time with 3 functions. It’ll be exactly the same but

with one extra step.

• Find (f ◦ g ◦ h)(x) given f, g, and h below.

f(x) = 2x (14)

g(x) = x

2 + 2x (15)

h(x) = 2x (16)

(17)

We wish to find f(g(h(x))). We will first find g(h(x)).

h(x) = 2x (18)

g( ) = ( )2 + 2( ) (19)

g(h(x)) = (h(x))2 + 2(h(x)) (20)

g(2x) = (2x)

2 + 2(2x) (21)

g(2x) = 4x

2 + 4x (22)

Thus g(h(x)) = 4x

2 + 4x. We now wish to find f(g(h(x))).

g(h(x)) = 4x

2 + 4x (23)

f( ) = 2( ) (24)

f(g(h(x))) = 2(g(h(x))) (25)

f(4x

2 + 4x) = 2(4x

2 + 4x) (26)

f(4x

2 + 4x) = 8x

2 + 8x (27)

(28)

Thus (f ◦ g ◦ h)(x) = f(g(h(x))) = 8x

2 + 8x.

4 0
3 years ago
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