a number, half of that number, and one-third of that number are added. the result is 22. what is the number?

1 answer:

5 0

Let the number be x.

Half of that number = x / 2
One third of the number = x / 3

= x / 2 + x / 3 = 22. LCM = 6.

= (3x + 2 x) / 6 = 22.

5x / 6 = 22. Cross multiply

5x = 6 * 22.

5x = 132 Divide both sides by 5.

x = 132 / 5

x = 26.4

The number is 26.4. Cheers.

You might be interested in

Write a polynomial function in factored form with zeros at −2, 5, and 6.

Anna007 [38]

The root -2 leads to the factor (x+2). The other 2 factors are (x-5) and (x-6).

Then the poly is f(x) = (x+2)(x-5)(x-6).

3 0

3 years ago

I think it is impossible. can someone please help with this one?

Maksim231197 [3]

That does look impossible someone plz help this man out or girl I don’t wanna assume genders ya know

8 0

2 years ago

Which of the following is a solution of z^5 = 1 + √3 i?

nordsb [41]

Answer:

Option 2 is right

Step-by-step explanation:

Given that

$z^5=1+\sqrt{3} i$

We can write this in polar form with modulus and radius

$|z^5|= \sqrt{1+3} =2\\tan of Angle t =\sqrt{3} \\$

Hence angle = 60 degrees and

$|z^5|= 2(cos60+isin60)$

Since we have got 5 roots for z, we can write 60, 420, 780, etc. with periods of 360

Using Demoivre theorem we get 5th root would be

5th root of 2 multiplied by 1/5 th of 60, 420, 780,....

$z= \sqrt[5]{2} (cos12+isin12)\\z=\sqrt[5]{2} (cos84+isin84)\\\\z=\sqrt[5]{2} (cos156+isin156)\\\\z=\sqrt[5]{2} (cos228+isin228)\\\\z=\sqrt[5]{2} (cos300+isin300)\\$