All categories

4 years ago

Identify the factors of x2 + 25y2. ...?

1 answer:

4 0

Is the equation like this? x^2 + 25y^2?
If the equation is like that, the answer would be (x-5y)^2 or (x-5y)(x-5y).
how did I get this? Just divide the factors or the get the square root of 25y^2 and it would be 5y. Then just try if it is positive or negative. In this case, it is negative because there is no factor w/ no exponential form. I hopemy answer helped you.

You might be interested in

A student took a system of equations, multiplied the first equation by 2 and the second equation by 3, then added the results to

egoroff_w [7]

Answer:

system 4.

Step-by-step explanation:

3x+6y=9

multiply by 2

6x+12y=18 ....(1)

-2x+-4y=4

multiply by 3

-6x-12y=12 ...(2)

add (1) and (2)

0+0=30

which is impossible .It has no solutions.

3 0

3 years ago

An arithmetic sequence is given below.

Mashcka [7]

Given that $a_1=24$ and $a_2=17$ , if $a_n$ is an arithmetic sequence, then the common difference between successive terms is $d=a_2-a_1=17-24=-7$ .

You then have

$a_2=a_1+d$
$\implies a_3=a_2+d=a_1+2d$
$\implies a_4=a_3+d=a_1+3d$
$\implies \cdots\implies a_n=a_{n-1}+d=\cdots=a_1+(n-1)d$

So the explicit formula for the $n$ th term is

$a_n=24-7(n-1)$

3 0

3 years ago

Find A U B U C. A){2,3,5,6,7,8,9,11,12}B){2,3,5,6,7,8,9}C){2,3,5,7,8,9,12}D){3,5,6,7,8,11,12}

Triss [41]

Answer: {2,3,5,6,7,8,9,11,12}

4 0

3 years ago

A bag contains 30 lottery balls numbered 1-30. A ball is selected, replaced, then another is drawn. Find each probability.

swat32

Answer:

17÷3 ok so it was just a compliment sometime soon

3 0

2 years ago

What is the number of subsets of S= {1, 2, 3…10} that contain five element include 3 or 4 but not both?

Ivenika [448]

Answer:

140

Step-by-step explanation:

To construct a subset of S with said property, we have two choices, include 3 in the subset or include four in the subset. These events are mutually exclusive because 3 and 4 can not both be elements of the subset.

First, let's count the number of subsets that contain the element 3.

Any of such subsets has five elements, but since 3 is already an element, we only have to select four elements to complete it. The four elements must be different from 3 and 4 (3 cannot be selected twice and the condition does not allow to select 4), so there are eight elements to select from. The number of ways of doing this is ${}_8C_4=70$ .

Now, let's count the number of subsets that contain the element 4.

4 is already an element thus we have to select other four elements . The four elements must be different from 3 and 4 (4 cannot be selected twice and the condition does not allow to select 3), so there are eight elements to select from, so this can be done in ${}_8C_4=70$ ways.

We conclude that there are 70+70=140 required subsets of S.

7 0

3 years ago