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bazaltina [42]
3 years ago
5

Find the range of the following piecewise function.

Mathematics
1 answer:
Anna11 [10]3 years ago
5 0

Given:

The piecewise function is

f(x)=\begin{cases}x+4 & \text{ if } -4\leq x

To find:

The range of given piecewise function.

Solution:

Range is the set of output values.

Both functions f(x)=x+4 and f(x)=2x-1 as linear functions.

Starting value of f(x)=x+4 is at x=-4 and end value is at x=3.

Starting value: f(-4)=-4+4=0

End value: f(3)=3+4=7

Starting value of f(x)=2x-1 is at x=3 and end value is at x=6.

Starting value: f(3)=2(3)-1=5

End value: f(6)=2(6)-1=11

Least range value is 0 at x=-4 and 0 is included in the range because -4 is included in the domain.

Largest range value is 11 at x=6 and 11 is not included in the range because 6 is not included in the domain.

So, the range of the given piecewise function is [0,11).

Therefore, the correct option is A.

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Answer:

10 and 15

Step-by-step explanation:

let Cairo's age be x then Poppy's age is x + 5 , then

x + x + 5 = 25

2x + 5 = 25 ( subtract 5 from both sides )

2x = 20 ( divide both sides by 2 )

x = 10

That is Cairo is 10 years old and Poppy is 15

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Add 110% of the cost to the cost.

Alternatively, multiply the cost by 2.1.
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Leon verified that the side lengths 21, 28, 35 form a Pythagorean triple using this procedure. Step 1: Find the greatest common
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Answer:

Yes, multiplying every length of a Pythagorean triple by the same whole number results in a Pythagorean triple.

Step-by-step explanation:

The missing options are:

<em>Yes, multiplying every length of a Pythagorean triple by the same whole number results in a Pythagorean triple. </em>

<em>Yes, any set of lengths with a common factor is a Pythagorean triple. </em>

<em>No, the lengths of Pythagorean triples cannot have any common factors. </em>

<em>No, the given side lengths can form a Pythagorean triple even if the lengths found in step 2 do not.</em>

A Pythagorean triple is a group of three integers (a, b and c) that satisfies the next equation:

c² = a² + b²

Multiplying the three integers by the same positive integer, you would get another Pythagorean triple.

(kc)² = (ka)² + (kb)²

k²c² = k²a² + k²b²

k²c² = k²(a² + b²)

c² = a² + b²

The procedure followed by Leon is the opposite. He found the greatest common factor and divided the given lengths by the greatest common factor, obtaining the simplest Pythagorean triple, which is (3,4,5)

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A container has red blue and white marbles. Four times the number of white marbles exceeded 9 times the number of red marbles by
RSB [31]
First we need to get some expressions:
"Four times the number of white marbles" = 4*w
"9 times the number of red marbles" = 9*r
"...exceeded...by 10": So 4*w is 10 more than 9*r or 4*w = 9*r +10
"ratio of blue marbles to red marbles is 3 to 1" : b/r = 3/1
b=3*r
"there is a total of 65 marbles": b+r+w=65
So our system becomes

b = 3*r
65 = b+r+w
4*w = 9*r+10

The first equation allows us to substitute immediately into the second equation so
65 = 4*r+w
however, we can divide the third equation by 4 to get w:
w = (9*r)/4+10/4
This can then be plugged into the second to get:
65 = 4*r+(9*r)/4+10/4 = (16*r)/4 + (9*r)/4+10/4 = (25*r+10)/4
260 = 25*r+10
250 = 25*r
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Now we simply plug this result into the substitutions to get w and b:
w = 90/4+10/4 = 100/4 = 25
b=30

So there are 10 red marbles, 25 white marbles, and 30 blue marbles

Now it sounds like we should be picking some out at random ;)

3 0
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