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3 years ago

Approximate square root of 39 to the nearest whole number

Mathematics

1 answer:

Dominik [7]3 years ago

5 0

It is about 6.24 so it rounds to 6

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PLEASE HELP EASY MATH

Alexxx [7]

Answer:90

Step-by-step explanation:

When the sum of two angles is 90°, then the angles are known as complementary angles. In other words, if two angles add up to form a right angle, then these angles are referred to as complementary angles.

7 0

3 years ago

Li deposited $17,500 into a bank account that earned simple interest each year. After 2 years, he had earned $2975 in interest.

Nimfa-mama [501]

Let x is the interest rate

$17,500 * 2 * x = <span>$2975
</span>$35,000x = $2,975
x = $2,975/$35,000
<span>x = 0.085

0.085 * 100 = 8.5%

Answer: annual interest rate was 8.5%

</span>

3 0

3 years ago

Read 2 more answers

Please help me 13 points

Fittoniya [83]

A. 28
B. 67.45
C. 71.4
Hope this helps sorry if I’m wrong

6 0

3 years ago

If x+3/3 = y+2/2, then x/3 = ______

timurjin [86]

X+1=y+1

X-Y= 0

Hope this helped you

7 0

2 years ago

Prove the following DeMorgan's laws: if LaTeX: XX, LaTeX: AA and LaTeX: BB are sets and LaTeX: \{A_i: i\in I\} {Ai:i∈I} is a fam

MariettaO [177]

$X-(A\cup B)=(X-A)\cap(X-B)$

I'll assume the usual definition of set difference, $X-A=\{x\in X,x\not\in A\}$ .

Let $x\in X-(A\cup B)$ . Then $x\in X$ and $x\not\in(A\cup B)$ . If $x\not\in(A\cup B)$ , then $x\not\in A$ and $x\not\in B$ . This means $x\in X,x\not\in A$ and $x\in X,x\not\in B$ , so it follows that $x\in(X-A)\cap(X-B)$ . Hence $X-(A\cup B)\subset(X-A)\cap(X-B)$ .

Now let $x\in(X-A)\cap(X-B)$ . Then $x\in X-A$ and $x\in X-B$ . By definition of set difference, $x\in X,x\not\in A$ and $x\in X,x\not\in B$ . Since $x\not A,x\not\in B$ , we have $x\not\in(A\cup B)$ , and so $x\in X-(A\cup B)$ . Hence $(X-A)\cap(X-B)\subset X-(A\cup B)$ .

The two sets are subsets of one another, so they must be equal.

$X-\left(\bigcup\limits_{i\in I}A_i\right)=\bigcap\limits_{i\in I}(X-A_i)$

The proof of this is the same as above, you just have to indicate that membership, of lack thereof, holds for all indices $i\in I$ .

Proof of one direction for example:

Let $x\in X-\left(\bigcup\limits_{i\in I}A_i\right)$ . Then $x\in X$ and $x\not\in\bigcup\limits_{i\in I}A_i$ , which in turn means $x\not\in A_i$ for all $i\in I$ . This means $x\in X,x\not\in A_{i_1}$ , and $x\in X,x\not\in A_{i_2}$ , and so on, where $\{i_1,i_2,\ldots\}\subset I$ , for all $i\in I$ . This means $x\in X-A_{i_1}$ , and $x\in X-A_{i_2}$ , and so on, so $x\in\bigcap\limits_{i\in I}(X-A_i)$ . Hence $X-\left(\bigcup\limits_{i\in I}A_i\right)\subset\bigcap\limits_{i\in I}(X-A_i)$ .

4 0

3 years ago