Learn Algebraic Expressions And Identities with Free Lessons & Tips

To find the value of x2−15x2−51 at x=−1x=−1, substitute x=−1x=−1 into the expression:

(−1)2−15(−1)2−51

1−151−51

To combine the terms with a common denominator, express 1 as 5555:

55−1555−51

4554

So, the value of x2−15x2−51 at x=−1x=−1 is 4554.

To find the value of x2+y2−10x2+y2−10 at x=0x=0 and y=0y=0, substitute these values into the expression:

(0)2+(0)2−10(0)2+(0)2−10

0+0−100+0−10

−10−10

So, the value of x2+y2−10x2+y2−10 at x=0x=0 and y=0y=0 is −10−10.

To find the product of 9a9a, 4ab4ab, and −2a−2a, simply multiply the coefficients (numbers) and the variables together:

(9a)×(4ab)×(−2a)(9a)×(4ab)×(−2a)

First, multiply the coefficients:

9×4×(−2)=−729×4×(−2)=−72

Now, multiply the variables:

a×a×b×a=a3ba×a×b×a=a3b

Combine the results:

(−72)×a3b(−72)×a3b

So, the product of 9a9a, 4ab4ab, and −2a−2a is −72a3b−72a3b.

To simplify the expression (a+b+c)(a+b−c)(a+b+c)(a+b−c), you can use the distributive property (FOIL method):

(a+b+c)(a+b−c)(a+b+c)(a+b−c)

=a(a+b−c)+b(a+b−c)+c(a+b−c)=a(a+b−c)+b(a+b−c)+c(a+b−c)

Now, distribute each term:

=a2+ab−ac+ba+b2−bc+ca+cb−c2=a2+ab−ac+ba+b2−bc+ca+cb−c2

Combine like terms:

=a2+2ab−ac+b2−bc−c2=a2+2ab−ac+b2−bc−c2

So, the simplified form of (a+b+c)(a+b−c)(a+b+c)(a+b−c) is a2+2ab−ac+b2−bc−c2a2+2ab−ac+b2−bc−c2.

To evaluate the product 8.56×11.608.56×11.60, you can simply multiply the two numbers:

8.56×11.60=99.5368.56×11.60=99.536

So, 8.56×11.608.56×11.60 is equal to 99.53699.536.

To evaluate (99)2(99)2, simply square the number 99:

(99)2=99×99(99)2=99×99

(99)2=9801(99)2=9801

Therefore, (99)2(99)2 is equal to 9801.

To simplify the expression x(2x−1)+5x(2x−1)+5 and find its value at x=−2x=−2, follow these steps:

x(2x−1)+5x(2x−1)+5

Distribute the xx into the parentheses:

2x2−x+52x2−x+5

Now, substitute x=−2x=−2 into the expression:

2(−2)2−(−2)+52(−2)2−(−2)+5

2(4)+2+52(4)+2+5

8+2+58+2+5

1515

So, the simplified expression is 2x2−x+52x2−x+5, and its value at x=−2x=−2 is 1515.

To evaluate (95)2(95)2 using identities, you can use the square of a binomial formula:

(a+b)2=a2+2ab+b2(a+b)2=a2+2ab+b2

In this case, a=90a=90 and b=5b=5. Apply the formula:

(95)2=(90+5)2(95)2=(90+5)2

=902+2×90×5+52=902+2×90×5+52

=8100+900+25=8100+900+25

=9025=9025

So, (95)2(95)2 is equal to

To calculate the volume of a cuboidal box, you multiply its three dimensions (length, width, and height). In this case, the dimensions are given as 5x5x, 3x23x2, and 7x47x4:

Volume=Length×Width×HeightVolume=Length×Width×Height

Volume=(5x)×(3x2)×(7x4)Volume=(5x)×(3x2)×(7x4)

Now, multiply the terms:

Volume=105x1+2+4Volume=105x1+2+4

Combine the exponents:

Volume=105x7Volume=105x7

So, the volume of the cuboidal box is 105x7105x7.

A histogram is used to represent the distribution of continuous data, typically grouped into intervals or bins. Let's analyze the given scenarios:

(i) The number of letters for different areas in a postman’s bag:

• This scenario involves discrete data, as the number of letters cannot be measured in continuous intervals. A bar graph or a simple bar chart may be more appropriate for displaying this type of data, where each bar represents the count of letters for a specific area.

(ii) The height of competitors in an athletics meet:

• This scenario involves continuous data, as height can vary within a range and is measured on a continuous scale. A histogram would be suitable for representing the distribution of heights, as it allows for the visualization of the frequency distribution across different height intervals.

Therefore, for the height of competitors in an athletics meet, a histogram would be more appropriate to show the distribution of heights.