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krek1111 [17]
3 years ago
10

What are the zeros of this function?

Mathematics
1 answer:
ElenaW [278]3 years ago
4 0

Answer:

0, and 4

Step-by-step explanation:

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The universal set in this diagram is the set of integers from 1 to 15. place the integers in the correct place in the venn diagr
neonofarm [45]

Answer:


Step-by-step explanation:

Given :  universal set in this diagram is the set of integers from 1 to 15.

Solution :

The intersection of odd integer,multiples of 3 and Factors of 15 are 3,15

The intersection of odd integer and Factors of 15 are 1,5

The intersection of odd integer,multiples of 3 is 9

The remaining multiples of 3 are 6,12

The remaining odd integers are 7,11,13

Now the remaining integers are 2,4,8,10,14 and these integers must be placed in the boxes outside the circles Since they does not belong any intersection or odd integer or factor of 15 .

Refer the attached figure for the answer.

3 0
3 years ago
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How do you do 4x - 9 = 6x - 13? what do you subtract from both sides?
ohaa [14]
To start, you need to subtract 4x from both sides, that way, x is one one sid of the equation.

8 0
4 years ago
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Use the distributive property to find a equivalent expression for 6(x+4)
MArishka [77]

Answer:

6x +24

Step-by-step explanation:

you factor out the 6 so 6x and 6*4 is 24

5 0
3 years ago
Greatest common factor of three univariate monomials
ryzh [129]

Answer:

3

Step-by-step explanation:

The GCF is the largest number that is a factor of the numbers listed. In this case, since one of the numbers is a constant the GCF will not have a variable. However, all of the numbers can be divided by 3. This is because 3 is the GCF.

3 0
3 years ago
Consider all 5 letter "words" made from the full English alphabet. (a) How many of these words are there total? (b) How many of
VARVARA [1.3K]

Answer:

a) There are 11,881,336 of these words in total.

b) There are 7,893,600 of these words with no repeated letters.

c) 896,376 of these words start with an a or end with a z or both

Step-by-step explanation:

Our words have the following format:

L1 - L2 - L3 - L4 - L5

In which L1 is the first letter, L2 the second letter, etc...

There are 26 letters in the English alphabet.

(a) How many of these words are there total?

Each of L1, L2, L3, L4 and L5 have 26 possible options.

So there are 26^{5} = 11,881,336 of these words total

(b) How many of these words contain no repeated letters?

The first letter can be any of them, so L1 = 26.

At the second letter, the first one cannot be repeated, so L2 = L1 - 1 = 25.

At the third letter, nor the first nor the second one can be repeated, so L3 = L1 - 2 = 24

This logic applies until L5

So we have

26-25-24-23-22

In total there are

26*25*24*23*22 = 7,893,600

of these words with no repeated letters.

(c) How many of these words start with an a or end with a z or both (repeated letters are allowed)?

T = T_{1} + T_{2} + T_{3}

T_{1} is the number of words that start with an a and do not end with z. So we have

1 - 26 - 26 - 26 - 25

The first letter can only be a, and the last one cannot be z. So:

T_{1} = 26^{3}*25 = 439,400

T_{2} is the number of words that start with any letter other than a and end with z. So we have

25 - 26 - 26 - 25 - 1

The first letter can be any of them, other than a, and the last can only be z. So:

T_{2} = 26^{3}*25 = 439,400

T_{3} is the number of words that both start with a and end with z. So:

1 - 26 - 26 - 26 - 1

The first letter can only be a, and the last can only be z. The other three letters could be anything. So:

T_{3} = 26^{3} = 17,576

T = T_{1} + T_{2} + T_{3} = 2*439,400 + 17,576 = 896,376

896,376 of these words start with an a or end with a z or both

4 0
3 years ago
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