All categories

3 years ago

Find the smallest integer k such that 60k is aperfect square

Mathematics

1 answer:

Travka [436]3 years ago

5 0

Answer: k = 15

Step-by-step explanation:

60 times some number k must be a perfect square.

Factor 60 to see which values do not have a pair:

60 = 2 · 2 · 3 · 5 --> 60 = 2 · 2 · 3 · 5

↓

3 · 5 are not pairs

So, you need to multiply 60 by 15 to make a perfect square

60 × 15 = 900 → $\sqrt{900}=30$

You might be interested in

Question 3, help please!

White raven [17]

The answer is not defined.

Explanation:

The given matrix is $\left[\begin{array}{cc}{2} & {4} \\ {1} & {-6}\end{array}\right]+\left[\begin{array}{c}{1} \\ {0}\end{array}\right]$

The matrix $\left[\begin{array}{cc}{2} & {4} \\ {1} & {-6}\end{array}\right]$ has dimensions $2\times 2$

This means that the matrix has 2 rows and 2 columns.

Also, the matrix $\left[\begin{array}{l}{1} \\ {0}\end{array}\right]$ has dimensions $2\times1$

This means that the matrix has 2 rows and 1 column.

Since, the matrices can be added only if they have the same dimensions.

In other words, to add the matrices, the two matrices must have the same number of rows and same number of columns.

Since, the dimensions of the two matrices are not equal, the addition of these two matrices is not possible.

Hence, the addition of these two matrices is not defined.

4 0

2 years ago

<img src="https://tex.z-dn.net/?f=%28w%20%2B%20%20%5Cfrac%7B4%7D%7Bw%7D%20%29%20%20%7B%7D%5E%7B2%7D%20" id="TexFormula1" title="

kifflom [539]

Answer:

W^2 + 8w/w + 16/x^2

Explanation:

6 0

2 years ago

7 Find the value of the expression. Z + 2y for y = 6 and z = 1

Volgvan

Answer:

Step-by-step explanation:

1+2×6=13

boy!!!!!!!!!!!!!!

8 0

2 years ago

Which BEST describes the difference between the medians of boys and girls as a multiple of the interquartile range for the boys.

rusak2 [61]

Answer:

B) 7\8

Step-by-step explanation:

Boy interquartile range = 4

Difference between medians = 3.5

therefore,

3.5 \4 = 7\8

5 0

2 years ago

Find a particular solution to <img src="https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%20%5Cfrac%7B%20d%5E%7B2%7Dy%20%7D%7Bd%20x%5E%7

Digiron [165]

$y=x^r$
$\implies r(r-1)x^r+6rx^r+4x^r=0$
$\implies r^2+5r+4=(r+1)(r+4)=0$
$\implies r=-1,r=-4$

so the characteristic solution is

$y_c=\dfrac{C_1}x+\dfrac{C_2}{x^4}$

As a guess for the particular solution, let's back up a bit. The reason the choice of $y=x^r$ works for the characteristic solution is that, in the background, we're employing the substitution $t=\ln x$ , so that $y(x)$ is getting replaced with a new function $z(t)$ . Differentiating yields

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dz}{\mathrm dt}$
$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac1{x^2}\left(\dfrac{\mathrm d^2z}{\mathrm dt^2}-\dfrac{\mathrm dz}{\mathrm dt}\right)$

Now the ODE in terms of $t$ is linear with constant coefficients, since the coefficients $x^2$ and $x$ will cancel, resulting in the ODE

$\dfrac{\mathrm d^2z}{\mathrm dt^2}+5\dfrac{\mathrm dz}{\mathrm dt}+4z=e^{2t}\sin e^t$

Of coursesin, the characteristic equation will be $r^2+6r+4=0$ , which leads to solutions $C_1e^{-t}+C_2e^{-4t}=C_1x^{-1}+C_2x^{-4}$ , as before.

Now that we have two linearly independent solutions, we can easily find more via variation of parameters. If $z_1,z_2$ are the solutions to the characteristic equation of the ODE in terms of $z$ , then we can find another of the form $z_p=u_1z_1+u_2z_2$ where

$u_1=-\displaystyle\int\frac{z_2e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$
$u_2=\displaystyle\int\frac{z_1e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$

where $W(z_1,z_2)$ is the Wronskian of the two characteristic solutions. We have

$u_1=-\displaystyle\int\frac{e^{-2t}\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_1=\dfrac23(1-2e^{2t})\cos e^t+\dfrac23e^t\sin e^t$

$u_2=\displaystyle\int\frac{e^t\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_2=\dfrac13(120-20e^{2t}+e^{4t})e^t\cos e^t-\dfrac13(120-60e^{2t}+5e^{4t})\sin e^t$

$\implies z_p=u_1z_1+u_2z_2$
$\implies z_p=(40e^{-4t}-6)e^{-t}\cos e^t-(1-20e^{-2t}+40e^{-4t})\sin e^t$

and recalling that $t=\ln x\iff e^t=x$ , we have

$\implies y_p=\left(\dfrac{40}{x^3}-\dfrac6x\right)\cos x-\left(1-\dfrac{20}{x^2}+\dfrac{40}{x^4}\right)\sin x$

4 0

2 years ago

Find the smallest integer k such that 60k is aperfect square​

Find the smallest integer k such that 60k is aperfect square