How to find the smallest integer k such that 60k is a perfect square

Mathematics

1 answer:

suter [353]3 years ago

6 0

9514 1404 393

Answer:

k = 15

Step-by-step explanation:

Look for the missing factors that make all of the factors of 60k be squares.

60 = 2² × 3 × 5

The factors 3 and 5 each have an odd exponent (1), so those two factors must be part of k.

60k = 2² × 3 × 5 × k

is a perfect square when ...

3 × 5 = k = 15

For k = 15, 60k = 900 = 30²

You might be interested in

PLEASE HELP ASAP WILL MARK AS BRAINLYEST

Stolb23 [73]

Answer:

The answer is 40.

Step-by-step explanation:

4x5 for the square part.

2x10 for the rectangle part

add them

5 0

3 years ago

Read 2 more answers

The midpoint of AB is at (3,7) and A is at (0,-5). Where is B located?

olga nikolaevna [1]

$\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ \square }}\quad ,&{{ \square }})\quad 
% (c,d)
&({{ \square }}\quad ,&{{ \square }})
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)\qquad thus
\\
----------------------------\\$ $\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ 0}}\quad ,&{{ -5}})\quad 
% (c,d)
B&({{ \square }}\quad ,&{{ \square }})
\end{array}\qquad
% coordinates of midpoint 
(3,7)\impliedby midpoint\qquad thus
\\ \quad \\
\left(\cfrac{{{ x_2 }} + {{ 0}}}{2}=3\quad ,\quad \cfrac{{{ y_2 }} + {{( -5)}}}{2}=7 \right)\to 
\begin{cases}
\cfrac{{{ x_2 }} + {{ 0}}}{2}=3
\\ \quad \\
\cfrac{{{ y_2 }} + {{ -5}}}{2}=7
\end{cases}
\\ \quad \\
solve\ for\ x_2\ and\ y_2$