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3 years ago

Is pi over 5 a whole number

Mathematics

2 answers:

dezoksy [38]3 years ago

4 0

Answer: Nooo!!!!

Step-by-step explanation:

Pi is not a whole number!!! It is a constantly but never repeating decimal. Pi divides by 5 is not a whole number. Whole numbers are evenly divisible.

Masja [62]3 years ago

3 0

Answer:

yes i think so

Step-by-step explanation:

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Which are counterexamples?<br> Look at the picture below. <br> Choose all that apply.

Flauer [41]

Answer:

$x^4/(2x^6)$

Explanation:

In the first example the division results in a radical, not a polynomial.

The remaining examples are not counter-examples (they do result in a polynomial)

8 0

3 years ago

Solve the following question algebraically. SOMEONE PLEASE HELP ME! ILL MARK THE BRAINLIEST ANSWER!!I’VE TRIED TO FIGURE IT OUT

Delicious77 [7]

Solve for x. Isolate the x. Note the equal sign. What you do to one side, you do to the other. Do the opposite of PEMDAS.

First, multiply 3 to both sides

6(3) = ((x + 2)/3)(3)

18 = x + 2

Finally, isolate the x. Subtract 2 from both sides

18 (-2) = x + 2 (-2)

x = 18 - 2

x = 16

16 is your answer for x

hope this helps

8 0

3 years ago

The set of all whole numbers not greater than 5

k0ka [10]

Answer:

-5, -4 , -3 , -2, -1, 1, 2 , 3, 4, 5

Step-by-step explanation:

6 0

3 years ago

(6-3(cube root of 6)/(cube root of 9)

MatroZZZ [7]

For this case we must simplify the following expression:

$\frac {6-3 \sqrt [3] {6}} {\sqrt [3] {9}}$

Multiplying the numerator and denominator by $(\sqrt [3] {9}) ^ 2$

$\frac {6-3 \sqrt [3] {6}} {\sqrt [3] {9}} * \frac {(\sqrt [3] {9}) ^ 2} {(\sqrt [3] { 9}) ^ 2} =$

We rewrite:

$\frac {\frac {6-3 \sqrt [3] {6}} * (\sqrt [3] {9}) ^ 2} {\sqrt [3] {9} * (\sqrt [3] {9 }) ^ 2} =$

By properties of powers we have that:

$a ^ m * a ^ n = a ^ {m + n}\\\frac {(6-3 \sqrt [3] {6}) * (\sqrt [3] {9}) ^ 2} {(\sqrt [3] {9}) ^ 3} =\\\frac {(6-3 \sqrt [3] {6}) * (\sqrt [3] {9}) ^ 2} {9} =$

We rewrite, moving the exponent within the radical:

$\frac {(6-3 \sqrt [3] {6}) * \sqrt [3] {9 ^ 2}} {9} =\\\frac {(6-3 \sqrt [3] {6}) * \sqrt [3] {81}} {9} =$

We can rewrite $3 * 3 ^ 3 = 81$

$\frac {(6-3 \sqrt [3] {6}) * \sqrt [3] {3 * 3 ^ 3}} {9} =$

We simplify:

$\frac {(6-3 \sqrt [3] {6}) * 3 \sqrt [3] {3}} {9} =$

We apply distributive property:

$\frac {18 \sqrt [3] {3} -9 \sqrt [3] {18}} {9} =$

Simplifying we finally have:

$2 \sqrt [3] {3} - \sqrt [3] {18}$

Answer:

$2 \sqrt [3] {3} - \sqrt [3] {18}$

5 0

3 years ago

The value of a dirt bike decreases by 20% each year. If you purchased this dirt bike today for $500, to the nearest dollar how m

NeTakaya

The bike is now $500. The price goes down by 20% each year.

500(0.20) = 100

The bike decreases by $100 each year.

After 5 years at $100 less per year, the bike will be $0.00.

4 0

3 years ago