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AURORKA [14]
4 years ago
14

What is x*x=15? (Basically make an equation that equals up to 15.)

Mathematics
1 answer:
rodikova [14]4 years ago
3 0

Answer:

\displaystyle \sqrt{15} = x

Step-by-step explanation:

Be VERY careful. When it write out the equation that way, it will give the speculation of this:

\displaystyle 15 = x^2

Now, in this case, since you were extra explanatory about this [what is written in parentheses], two factors I know of that multiply to 15 are these:

** \displaystyle 15 = 2 \times 7\frac{1}{2}

I am joyous to assist you at any time. ☺️

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arlik [135]

Answer:

1/2

Step-by-step explanation:

from one point to another it goes up 1 and over two. so rise over run, 1 over 2

7 0
3 years ago
Read 2 more answers
SOMEONE ANSWER QUICK!!
aleksandrvk [35]

Answer:

3/4 of a pizza leftover.

Step-by-step explanation:

2 7/12 + 2/3 = 3 1/4

4 -  3 1/4 = 3/4

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6 0
3 years ago
(a) How many integers from 1 through 1,000 are multiples of 2 or multiples of 9?
DENIUS [597]

Answer:

(a)There are 500 integers from 1 through 1,000 are multiples of 2.

There are 111 integers from 1 through 1,000 are multiples of 9.

(b)0.611

Step-by-step explanation:

Given the set of Integers from 1 through 1000

The least multiple of 2 is 2 and the highest multiple of 2 in the interval is 1000.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 2,4,6,... is an Arithmetic Progression,

where first term, a =2, common difference, d =2

Nth term of an A.P,

U_n= a+(n-1)d\\1000=2+2(n-1)\\1000=2+2n-2\\1000=2n\\n=500

  • There are 500 integers from 1 through 1,000 are multiples of 2.

Similarly,

Given the set of Integers from 1 through 1000

The least multiple of 9 is 9 and the highest multiple of 9 in the interval is 999.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 9,18,27,... is an Arithmetic Progression,

where first term, a =9, common difference, d =9

Nth term of an A.P,

U_n= a+(n-1)d\\999=9+9(n-1)\\999=9+9n-9\\999=9n\\n=111

  • There are 111 integers from 1 through 1,000 are multiples of 9.

(b)Probability that an integer selected at random is a multiple of 2 or a multiple of 9.

There are a total of 1000 numbers between from 1 through 1,000

n(S)=1000

n(Multiples of 2)=500

n(Multiples of 9)=111

Therefore:

P(\text{Multiples of 9}) \:OR\: P(\text{Multiples of 2})=\frac{111}{1000} + \frac{500}{1000} \\=\frac{611}{1000} \\=0.611

7 0
3 years ago
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iogann1982 [59]

Answer:

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Step-by-step explanation:

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3 years ago
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Is there any option to choose from ?

If so show them and I’ll probably be able to answer .
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