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serious [3.7K]
3 years ago
8

What is the square root of 75

Mathematics
2 answers:
Wewaii [24]3 years ago
7 0
Square root (75) =
Square root (3 * 25) =
5 * Square root (3)

V125BC [204]3 years ago
3 0
<span>8.66025403784 to be exact.

</span>
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The point (0, 4) is located
Doss [256]

Answer:

it would be on y-4.

and there is no x coordinate because it's 0

3 0
2 years ago
Which of the following statements is false?
kramer

Option D

The sum of two irrational numbers is always rational is false statement

<em><u>Solution:</u></em>

<h3><u>The sum of two rational numbers is always rational</u></h3>

"The sum of two rational numbers is rational."

So, adding two rationals is the same as adding two such fractions, which will result in another fraction of this same form since integers are closed under addition and multiplication. Thus, adding two rational numbers produces another rational number.

For example:

\frac{1}{4} + \frac{1}{5} = \frac{9}{20}

rational number + rational number = rational number

Hence this statement is true

<h3><u>The product of a non zero rational number and an irrational number is always irrational</u></h3>

If you multiply any irrational number by the rational number zero, the result will be zero, which is rational.

Any other situation, however, of a rational times an irrational will be irrational.

A better statement would be: "The product of a non-zero rational number and an irrational number is irrational."

So this statement is correct

<h3><u>The product of two rational numbers is always rational</u></h3>

The product of two rational numbers is always rational.

A number is said to be a rational number if it is of the form p/q,where p and q are integers and q ≠ 0

Any integer is a rational number because it can be written in p/q form.

Hence it is clear that product of two rational numbers is always rational.

So this statement is correct

<h3><u>The sum of two irrational numbers is always rational</u></h3>

"The sum of two irrational numbers is SOMETIMES irrational."

The sum of two irrational numbers, in some cases, will be irrational. However, if the irrational parts of the numbers have a zero sum (cancel each other out), the sum will be rational.

Thus this statement is false

5 0
3 years ago
In Juan's coin bank there are nickels (n),
alukav5142 [94]

Answer:

The answer is 8 quarters, 16 nickles, and 32 dimes

Step-by-step explanation:

  1. First, the equation would be (2/4d) x 5 +d x 10 +1/4d x 25= 600 (Don't put 6.00, that's dollars. We put 6 dollars into cents to make it easier.)
  2. Then we slowly start to solve: 10/4d + 10d + 25/4d = 600
  3. 10d+40d+25d/4 = 600 (Put all of that over 4, not first 25d)
  4. 75d= 600 x 4
  5. d=600 x 4/75 (same here put everything over 75)
  6. d=32
  7. So then n= 2 x 32/4 = 16
  8. Then n would be equal to : 1/4d= 1/4 x 32 = 8
  9. So your answer would be 8 quarters, 16 nickles, and 32 dimes.

Hope this helped :)

5 0
3 years ago
The radius of star A is 6.9203x10^5 km, which is 108.7 times the rafius of star B. what is the radius of the star B?
Kay [80]

Answer:

6366.4213

Step-by-step explanation: 6.9203x10^5 is 692,030. I divided 692,030 by 108.7 to get the answer. Anybody can correct any error I may have.

7 0
2 years ago
9 and 5/12 - 3 and 2/3
lana [24]
Recall to always convert the mixed fractions to "improper" fractions first,

\bf \stackrel{mixed}{9\frac{5}{12}}\implies \cfrac{9\cdot 12+5}{12}\implies \stackrel{improper}{\cfrac{113}{12}}&#10;\\\\\\&#10;\stackrel{mixed}{3\frac{2}{3}}\implies \cfrac{3\cdot 3+2}{3}\implies \stackrel{improper}{\cfrac{11}{3}}\\\\&#10;-------------------------------\\\\&#10;\cfrac{113}{12}-\cfrac{11}{3}\impliedby \textit{clearly our LCD is 12}\implies \cfrac{113-(4)11}{12}&#10;\\\\\\&#10;\cfrac{113-44}{12}\implies \cfrac{69}{12}\implies 5\frac{9}{12}
5 0
3 years ago
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