All categories

3 years ago

Which shows a difference of squares?

Mathematics

1 answer:

Juli2301 [7.4K]3 years ago

6 0

A difference of squares must be a binomial, so we want two terms. This rules out the third and fourth options.

If we look at the first two, we must check that both terms are perfect squares.

Look at the first one: $4x^2$ is a perfect square (it's $(2x)^2$ ), but $10y^2$ is not a perfect square, because 10 is not a perfect square.

On the other hand, if we look at the second option, $16y^4$ is the square of $4y^2$ , and $x^2$ is clearly the square of $x$ . So, this is the difference of two squares.

You might be interested in

The width of each of five continuous classes in a frequency distribution is 5 and the lower class-limit of the lowest class is 1

Helga [31]

Let x and y be the upper and lower class limit of frequency distribution.

Given, width of the class = 5

⇒ x-y= 5 …

Also, given lower class (y) = 10 On putting y = 10, we get

x – 10= 5 ⇒ x = 15 So, the upper class limit of the lowest class is 15

Hence, the upper class limit of the highest class

=(Number of continuous classes x Class width + Lower class limit of the lowest class)

= 5 x 5+10 = 25+10=35

Hence,’the upper class limit of the highest class is 35.

<em><u>Alternate Method</u></em><em><u>.</u></em>

After finding the upper class limit of the lowest class, the five continuous classes in a frequency distribution with width 5 are 10-15,15-20, 20-25, 25-30 and 30-35.

Thus, the highest class is 30-35..

<h3>Hence, the upper limit of this class is 35.</h3>

4 0

3 years ago

First pirson to type 1 i will see if i can give brainlist but idk how bot i will try

amm1812

Answer:

Step-by-step explanation:

you asked to type 1

3 0

3 years ago

The graphs below are both absolute value functions. The equation of the redgraph is f(x) = |xl. Which of these is the equation o

grigory [225]

ANSWER

$\Rightarrow y=\frac{1}{2}|x|$

EXPLANATION

We want to find the absolute value function for the line in blue.

The general form of an absolute value function is:

$y=a|x-h|+k$

where (h, k) = vertex

From the line, the vertex of the graph in blue is:

$(0,0)$

To find a, we have to pick a point (x, y) on the line and input it into the general function.

Let us pick (2, 1).

Therefore, we have:

$\begin{gathered} 1=a|2-0|+0 \\ 1=a|2|=2a \\ \Rightarrow a=\frac{1}{2} \end{gathered}$

Therefore, the absolute value function is:

$\begin{gathered} y=\frac{1}{2}|x-0|+0 \\ \Rightarrow y=\frac{1}{2}|x| \end{gathered}$

3 0

1 year ago

What scale factor was applied to the first rectangle to get the resulting image? The answer is a decimal. The big rectangle is 2

Juli2301 [7.4K]

Scale factor is always (new/original)

so it depends which rectangle came first. if it started as large and was dilated to the smaller one, then the ratio is (.5/2.5) which simplifies to 1/5 or 0.2.

3 0

3 years ago

How can the average rate of change be found using a discrete graph?

nekit [7.7K]

In general, the average rate of change of f (x) on the interval a, b is given by f(b) – f(a) / b – a. The average rate of alteration of a function, f (x) on an interval is well-defined to be the variance of the function values at the endpoints of the interim divided by the difference in the x values at the endpoints of the interval. this is also known as the difference quotient that tells how on average, the y values of a function are changing in connection to variations in the x values. A positive or negative rate of change is applicable which match up to an increase or decrease in the y value among the two data points. It is called zero rate of change when a quantity does not change over time.

5 0

3 years ago