All categories

3 years ago

List the first 11 perfect squares

Mathematics

1 answer:

vlabodo [156]3 years ago

6 0

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121

You might be interested in

I need helppp plzzz :((

marshall27 [118]

Answer:

u=8

3u+5u=64

8u=64

u=8

finish

3 0

3 years ago

Read 2 more answers

What would an 18% increase of 90 be?

madam [21]

Answer:

106.2

Step-by-step explanation:

A 18% increase of a number is the same as multiplying it by 1.18.

If we multiply 90 by 1.18 we get:

90 * 1.18

106.2

8 0

3 years ago

Read 2 more answers

For the following linear system, put the augmented coefficient matrix into reduced row-echelon form.

Anni [7]

Answer:

The reduced row-echelon form of the linear system is $\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&0\\0&0&0&1\end{array}\right]$

Step-by-step explanation:

We will solve the original system of linear equations by performing a sequence of the following elementary row operations on the augmented matrix:

Interchange two rows
Multiply one row by a nonzero number
Add a multiple of one row to a different row

To find the reduced row-echelon form of this augmented matrix

$\left[\begin{array}{cccc}2&3&-1&14\\1&2&1&4\\5&9&2&7\end{array}\right]$

You need to follow these steps:

Divide row 1 by 2 $\left(R_1=\frac{R_1}{2}\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\1&2&1&4\\5&9&2&7\end{array}\right]$

Subtract row 1 from row 2 $\left(R_2=R_2-R_1\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\5&9&2&7\end{array}\right]$

Subtract row 1 multiplied by 5 from row 3 $\left(R_3=R_3-\left(5\right)R_1\right)$

$\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right]$

Subtract row 2 multiplied by 3 from row 1 $\left(R_1=R_1-\left(3\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right]$

Subtract row 2 multiplied by 3 from row 3 $\left(R_3=R_3-\left(3\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&0&0&-19\end{array}\right]$

Multiply row 2 by 2 $\left(R_2=\left(2\right)R_2\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&-19\end{array}\right]$

Divide row 3 by −19 $\left(R_3=\frac{R_3}{-19}\right)$

$\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&1\end{array}\right]$

Subtract row 3 multiplied by 16 from row 1 $\left(R_1=R_1-\left(16\right)R_3\right)$

$\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&-6\\0&0&0&1\end{array}\right]$

Add row 3 multiplied by 6 to row 2 $\left(R_2=R_2+\left(6\right)R_3\right)$

$\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&0\\0&0&0&1\end{array}\right]$

8 0

3 years ago

The sum of three numbers is 97. The third number is 2 times the second. The second number is 9 less than the first. What are the

inn [45]

Answer:

The numbers are 26.5, 53 and 17.5.

Step-by-step explanation:

97=x+(x-9)+2x

97+9=4x

106/4=4x/4

x= 26.5

x-9 = 17.5

2x= 53

8 0

3 years ago

Write the monomial in standard form. name it's coefficient and identify its degree.

Hunter-Best [27]

Answer:

$Standard\ Form = {3n^2} m^{-2}$

Step-by-step explanation:

Given

$\frac{2}{3m^2n} * 4.5n^3$

Required

Write in Standard Form

To start with; the two monomials have to be multiplied together;

$\frac{2}{3m^2n} * 4.5n^3$

$Standard\ Form = \frac{2 * 4.5n^3}{3m^2n}$

Split the numerator and the denominator

$Standard\ Form = \frac{2 * 4.5 * n^3}{3 * m^2 * n}$

Multiply Like terms

$Standard\ Form = \frac{9 * n^3}{3 * m^2 * n}$

Divide 9 by 3 to give 3

$Standard\ Form = \frac{3 * n^3}{m^2 * n}$

Divide n³ by n to n²

$Standard\ Form = \frac{3 * n^2}{m^2 }$

Split fraction

$Standard\ Form = {3 * n^2} * \frac{1}{m^2 }$

From laws of indices;

$\frac{1}{a^n} = a^{-n}$

$Standard\ Form = {3 * n^2} * \frac{1}{m^2 }$ becomes

$Standard\ Form = {3 * n^2} * m^{-2}$

Multiply all together

$Standard\ Form = {3n^2} m^{-2}$

5 0

3 years ago