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vovangra [49]
3 years ago
5

What would be the be the answer to 3+11t

Mathematics
1 answer:
mash [69]3 years ago
5 0

Your teacher asked you to represent 3 more than 8 multiplied by a number T as an expression


(3+8)t = 11t

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HELP PLEASEEE!!!! I START SCHOOL TOMORROW
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4. The range is -2 ≤ y ≤ 4

5. The domain is {-2 , -1 , 0 , 3.5 , 4.2}

6. The relation is not a function

Step-by-step explanation:

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# Question 4:

- The range of the function is the values of all y (output) which is

 corresponding to the values of all x (input)

- The figure is a stare, the values of x is from -4 to 4

∴ The domain is -4 ≤ x ≤ 4

- The corresponding values of y are from -2 to 4

∴The range is -2 ≤ y ≤ 4

# Question 5:

- The table represent a relation between x and y

- x is the input of the relation

- y is the output of the relation

∵ The input of the relation is called the domain of it

∴ The domain of the relation is all values of x in the table

- the values of x are -2 , -1 , 0 , 3.5 , 4.2

∴ The domain is {-2 , -1 , 0 , 3.5 , 4.2}

# Question 6:

- The relation can be a function if each ordered pair has a

  unique value of x (input)

- That means every x (input) has only one y (output)

- Ex : The relation {(a , b) , (c , d) , (a , f)} is not a function because

 the input a has two output b and f

∵ The relation is {(-4 , 0) , (-3 , 0) , (-2 , 1) , (1 , -2) , (-3 , 4)}

∵ The x = -3 has two values of y 0 and 4

∴ The relation is not a function

7 0
3 years ago
Find the value of x.<br> 2x<br> 120°<br> 700<br> A. 55<br> B. 35<br> C. 25<br> D. 10
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Answer:

for this question you need to take the 12p and 700 and add then dived by 2x which then gives you your answer

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3 years ago
A certain positive integer has exactly 20 positive divisors. What is the smallest number of primes that could divide the integer
Alika [10]

As per my explanation to (b) above, the largest number of primes that could factor such a number is 4.

Note that  2,3,5 and 7   are the smallest primes, then use the reasoning from

(b) above. we are looking for four exponents, that, when 1 is added to each and all are multiplied together, would equal 20.

But no such integers  k, l, m, and n  exist such that (k + 1)(l + 1)(m + 1) (n + 1)  = 20 where k, l,m, and n ≥ 1 so this number, whatever it is, can't have 4 prime factors

Let's drop 7 out of the mix and suppose it has just 3 prime factors 2, 3, and 5 again we are looking for three exponents,  that, when 1 is added to each and all are multiplied together, would equal 20.  Put another way, we are looking for k, l and m ≥ 1   such that (k + 1)(l + 1)(m + 1) = 20

Note that the  only possibility here  is when we have 2 *2 *5  = 20 and the smallest possible product would be 2^(4 )* 3^(1) * 5^(1) =  2^4 * 3 * 5 = 240

Now......the only remaining possibility is  that this number is composed of the two smallest primes, 2 and 3,  and we are looking for  some k  and l ≥ 1 such that (k + 1)(l + 1) = 20 clearly, the only possibilities  are when k = 4 and l = 5, or vice-versa

So this number would factor as either 2^3 * 3^4   = 648  or 2^4 * 3^3 = 432 and both are > 240.

Learn more about positive integers at

brainly.com/question/1367050

#SPJ4

8 0
2 years ago
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