Algebra

1sr find the EXTerior Angle. 4

EXT ANG = 360 / 50

EXT Ang = 36/5 = 7.2

Hence

Int Ang = 180 - 7.2 = 172.8.

Since there are 50 sides there are 50 interior angles. at 172.8 each

So sum of interior angles is 50 X 172.8 = 8640 degrees.

NB

Remember the table.

Sum of interior angles.

3 sides = 180

4 sides= 360

5 sides = 540

et seq/

For each additional side add 180 degrees.

You have answered the question for yourself!!!! --- '50'.

2x + 3x = 5x

-2x + 3y = -7

3x + y = 16

simultaneous equations.

Elimination a letter. Arbitrarily select 'y'

Hence multiply the bottom eq'n by ' -3'

Hence

-2x + 3y = -7

-9x - 3y = -48

Add the eq'ns/

-11x = =55

x = -55/-11 = 5

When x = 5 , substitute into either eq'n for 'y'

-2(5) + 3y = -7

-10 + 3y = -7

3y = 3

y = 1

Hence answer in ( x,y) fpr, is ( 5,1).

3x + y = 5

There is no answer for 'x' ^ 'y' as there are two unknowns, and no comparable values.

However. it is a linear equation, and can be rearranged into the straight line form .

y = -3x + 5

Hence

Gradient/slope is '-3' and the y0intercept is '5'.

p = 2s + 5

Subtract '5' from both sides

2s = p - 5

Divide both sides by '2'

s = [p - 5] / 2

p + 2 = 2 - 3p

Subtract '2' from both sides

p = -30

Subtract 'p' from both sides

0 = -2p

Hence 2p = 0

It folllows p = o

-3p^(2)+ p^(2) = -2p^(2)

66 squared = 66^(2) = 66 x 66

By Long Multiplication.

66

x66

3960 ) 66 X 60)

• 396 )66 X 6 )

4356

===== The answer

10

**If you mean how many ways three 3-digit numbers can add up to 005, the answer depends on whether the order of the numbers matters to you: • There are 21 unique equations if the order matters (e.g., counting 001 + 002 + 002 as different from 002 + 002 + 001). • There are 5 basic combinations if the order does not matter (just looking at the groups of numbers used). Here are the complete lists for both ways of looking at it:

If switching the position of the numbers counts as a new equation, and 000 counts as a 3 digit number those 5 combinations expand into 21 variations: Using 0, 0, and 5: • 000 + 000 + 005 = 005 • 000 + 005 + 000 = 005 • 005 + 000 + 000 = 005 Using 0, 1, and 4: • 000 + 001 + 004 = 005 • 000 + 004 + 001 = 005 • 001 + 000 + 004 = 005 • 001 + 004 + 000 = 005 • 004 + 000 + 001 = 005 • 004 + 001 + 000 = 005 Using 0, 2, and 3: • 000 + 002 + 003 = 005 • 000 + 003 + 002 = 005 • 002 + 000 + 003 = 005 • 002 + 003 + 000 = 005 • 003 + 000 + 002 = 005 • 003 + 002 + 000 = 005 Using 1, 1, and 3: • 001 + 001 + 003 = 005 • 001 + 003 + 001 = 005 • 003 + 001 + 001 = 005 Using 1, 2, and 2: • 001 + 002 + 002 = 005 • 002 + 001 + 002 = 005 • 002 + 002 + 001 = 005**

SIMPLY AND DIRECTLY I ORGANIZED NUMBERS LOW TO HIGH. FOR UNDERSTANDING MADE EASY!

**If you mean how many ways three 3-digit numbers can add up to 005, the answer depends on whether the order of the numbers matters to you: • There are 21 unique equations if the order matters (e.g., counting 001 + 002 + 002 as different from 002 + 002 + 001). • There are 5 basic combinations if the order does not matter (just looking at the groups of numbers used). Here are the complete lists for both ways of looking at it:

If switching the position of the numbers counts as a new equation, and 000 counts as a 3 digit number those 5 combinations expand into 21 variations: Using 0, 0, and 5: • 000 + 000 + 005 = 005 • 000 + 005 + 000 = 005 • 005 + 000 + 000 = 005 Using 0, 1, and 4: • 000 + 001 + 004 = 005 • 000 + 004 + 001 = 005 • 001 + 000 + 004 = 005 • 001 + 004 + 000 = 005 • 004 + 000 + 001 = 005 • 004 + 001 + 000 = 005 Using 0, 2, and 3: • 000 + 002 + 003 = 005 • 000 + 003 + 002 = 005 • 002 + 000 + 003 = 005 • 002 + 003 + 000 = 005 • 003 + 000 + 002 = 005 • 003 + 002 + 000 = 005 Using 1, 1, and 3: • 001 + 001 + 003 = 005 • 001 + 003 + 001 = 005 • 003 + 001 + 001 = 005 Using 1, 2, and 2: • 001 + 002 + 002 = 005 • 002 + 001 + 002 = 005 • 002 + 002 + 001 = 005**

x^(2) = 201.64

find square root of both sides.

Hence

x = sqrt( 201.64(

x = 14.2

y = kx + 1

y^(2) = 8x

Square up the linear eq'n and substitute.

Hence

y^(2) = ( kx + 1)^(2)

(kx + 1)^(2) = 8x

Square out the brackets

(kx)^(2) + 2kx + 1 = 8x

(kx)^(2) + (2k-8)x + 1 = 0

Apply the discriminant

b^(2) -4ac = 0

(2k - 8)^(2) - 4(k)^(2)(1)

4k^(2) - 32k + 64 - 4k^(2) = 0

32k - 64 = 0

32k = 64

k = 2

=64

0

a^(3) + a^(2)

Factors to

a^(2) [ a^(2) + 1 ]

NB THe terms in the square brackets do NOT factor. Reason, Any ywo squared terms with a positive (+) between then does NOT factor

Think of a^(2) + 1 as a^(2) + 1^(2) .

A pair of squared terms with a negative between them DOES factor.

So if you initial expression was a^(4) - a^(2)

Then

a^(2) [ a^(2) - 1^(2)]

a^(2)[a - 1)(a + 1) Fully factored.

Note the two different signs.

To solve an inequality, first isolate the variable on one side of the inequality sign, similar to how you would solve an equation. This involves performing the same operations on both sides, such as adding, subtracting, multiplying, or dividing, while remembering that if you multiply or divide by a negative number, you must reverse the inequality sign. After isolating the variable, express the solution in interval notation or graph it on a number line to represent all possible values that satisfy the inequality.

All three sides are of differentn lengths. All three angles are of different values.

However, the sum of the interior angles is ALWAYS 180 degrees.

13 yards @36 inches per yard.

If x=1 (because if x is 2 or more the number is really weird) then the gif is 2097152.

33.3333..... ( Recurring to infinity

or

33 1/3

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sqrt(66) = sqrt(2 x 3 x 11) =

sqrt(2) X sqrt(3) X sqrt(11) or

2^(1/2) X 3^(1/2) X 11^(1/2) = 66^(1/2)

Since 2.3. & 11 are all prime number , their square roots are Irrational. Hence sqrt(66) is also irrational .

5

There are 48 years between 1964 to 2012. This is because 2012-1964=48.