pls help explain if u can

Answer:

a2 – b2 = (a – b)(a + b)

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

a2 + b2 = (a + b)2 – 2ab

(a – b)2 = a2 – 2ab + b2

(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

(a – b – c)2 = a2 + b2 + c2 – 2ab + 2bc – 2ca

(a + b)3 = a3 + 3a2b + 3ab2 + b3 ; (a + b)3 = a3 + b3 + 3ab(a + b)

(a – b)3 = a3 – 3a2b + 3ab2 – b3 = a3 – b3 – 3ab(a – b)

a3 – b3 = (a – b)(a2 + ab + b2)

a3 + b3 = (a + b)(a2 – ab + b2)

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

(a – b)4 = a4 – 4a3b + 6a2b2 – 4ab3 + b4

a4 – b4 = (a – b)(a + b)(a2 + b2)

a5 – b5 = (a – b)(a4 + a3b + a2b2 + ab3 + b4

Answer:

a2 – b2 = (a – b)(a + b)

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

a2 + b2 = (a + b)2 – 2ab

(a – b)2 = a2 – 2ab + b2

(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

(a – b – c)2 = a2 + b2 + c2 – 2ab + 2bc – 2ca

(a + b)3 = a3 + 3a2b + 3ab2 + b3 ; (a + b)3 = a3 + b3 + 3ab(a + b)

(a – b)3 = a3 – 3a2b + 3ab2 – b3 = a3 – b3 – 3ab(a – b)

a3 – b3 = (a – b)(a2 + ab + b2)

a3 + b3 = (a + b)(a2 – ab + b2)

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

(a – b)4 = a4 – 4a3b + 6a2b2 – 4ab3 + b4

a4 – b4 = (a – b)(a + b)(a2 + b2)

a5 – b5 = (a – b)(a4 + a3b + a2b2 + ab3 + b4)

Answer:

answer is - 1

Step-by-step explanation:

- x + y = 3

x = 4

- ( - 4 ) + y = 3

4 + y = 3. take 4 to the right

y = 3 - 4

y = - 1

I hope this answers your question.

Answer:

The essence of Mathematics lies in its freedom

Answer:

5n

Step-by-step explanation:

cost of one helmet = $5

let 'n' be the number of person that is n=1,2,3,4........

note that the value of n is always a natural number

therefore the expression = 5n

where 5 is the cost one helmet

n is the number of persons

Answer:

Hello,

Step-by-step explanation:

n! + (n-1)!

=(n-1)! * n+ (n-1)!

=(n-1)! * (n+1)

Answer:

aslo bien jjjjjjj

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

Values of y are decreasing where x is in the range (-4;02)

Answer:

The bigger angle is 144 degrees and the smaller one is 36 degrees

Answer: Third Choice. Thousandths

Step-by-step explanation:

Concept:

Here, we need to know the order and name of each place value.

Please refer to the attachment below for the specified names.

Solve:

STEP ONE: Orde and name each place

2 ⇒ One Thousands

4 ⇒ Hundreds

5 ⇒ Tens

6 ⇒ Ones

.

1 ⇒ Tenths

3 ⇒ Hundredths

8 ⇒ One Thousandths

7 ⇒ Ten Thousandths

STEP TWO: Find the number [8] in the number

As we can see from the list above, 8 is at the right of the decimal point, thus, the place value is Thousandths.

Hope this helps!! :)

Please let me know if you have any questions

Answer:

6, 7

Step-by-step explanation:

6 x 7 = 42

Answer:

The ratio of senior members to junior members at the company is 1 : 3

Step-by-step explanation:

The given average number of years of experience of the senior employees, x = 16 years

The given average number of years of experience of the junior employees, y = 7 years

Let a represent the number of senior employees in the company, and let b represent the number of junior workers in the company, we have;

(16·a + 4·b)/(a + b) = 7

∴ (16·a + 4·b) = (a + b) × 7 = 7·a + 7·b

16·a - 7·a = 7·b - 4·b

9·a = 3·b

a/b = 3/9 = 1/3

a/b = 1/3

The above equation, expressed as a ratio is, a : b = 1 : 3

Therefore;

The ratio of senior members to junior members at the company, a : b = 1 : 3.

Answer:

2x^2 + 9x - 4

18x^2 - 3x - 10

Step-by-step explanation:

use foil method

Answer:

Below

Step-by-step explanation:

To find the volume of a sphere you can use this formula!

     V = 4/3πr^3

Plugging in the values....

V = 4/3π(49.5)^3

V = 5.08047 × 10^5

V = 5.1 x 10^5 in^3

If you wanted to write this is standard form it would be

V = 508047 in^3

Hope this helps!

Answer:

The Constant is 9/2

Answer:

Step-by-step explanation:

Xét tam giác DAB có: P là trung điểm AD, M là trung điểm AB

=> MP là đường trung bình của tam giác DAB => MP//BD và MP=[tex]\frac{1}{2}[/tex]BD  (1)

Xét tam giác DBC có: N là trung điểm DC, Q là trung điểm BC

=> QN là đường trung bình của tam giác DBC => QN//BD và QN=[tex]\frac{1}{2}[/tex]BD  (2)

Từ (1) và (2) => vecto MP song song cùng chiều với vecto QN

và độ dài MP = độ dài QN = [tex]\frac{1}{2}[/tex]BD

=> vecto MP = vecto QN

Tương tự xét các tam giác DAC và tam giác ABC => vecto MQ = vecto PN

Answer:

Total = x+3 kg

Step-by-step explanation:

Green apples = x kg, Red apples = 3 kg

Answer:

A' (-3,12)

B' (9,6)

C' (-6,-6)

Answered by GAUTHMATH

Answer:

[tex]x = - 8[/tex]

Step-by-step explanation:

First,Step write the given equation.

[tex] \frac{x}{2} + 7 = 3[/tex]

In Algebra, we learn how to solve for a missing term called a variable. We must use inverse operations to undo terms to isolate our variable. Inverse operations means that we do the opposite of the given.

For example, say we have a equation

[tex]x + 2 = 3[/tex]

We would subtract 2 from both sides. E.g( remember the golden rule of algebra). What we do to one side, we do the other. This applies when we solving for a variable.

This means that

[tex]x = 1[/tex]

And if we substitute 1 in for x, it is indeed true.

[tex]1 + 2 = 3 = 3[/tex]

Now, back to the question.

[tex] \frac{x}{2} + 7 = 3[/tex]

First, we subtract 7 from both sides to get rid of the 7 on the left side. Also remeber to undo terms that aren't included in the variable when solving these problems.

So now we got

[tex] \frac{x}{2} = - 4[/tex]

Then we multiply 2 by both sides since that the opposite of division.

[tex] \frac{x}{2} \times 2 = x = -4 \times 2 = - 8[/tex]

So this means that

[tex]x = - 8[/tex]

If we plug this in, this is true.

[tex] \frac{ - 8}{2} + 7 = 3[/tex]

Answer:

15

Step-by-step explanation:

following PEMDAS, we don't have () so we move on to exponents

5^2 is equal to 5 x 5 which is 25

40-25=15

[tex]\huge\textsf{Hey there!}[/tex]

[tex]\large\textsf{40 - 5}\mathsf{^2}\\\\\\\large\textsf{5}\mathsf{^2}\\\large\textsf{= 5}\times\large\textsf{5}\\\large\textsf{= \bf 25}\\\\\\\large\textsf{= 40 - 25}\\\\\\\large\textsf{= \bf 15}\\\\\\\\\\\\\boxed{\boxed{\huge\textsf{Therefore your answer is: \bf 15}}}\huge\checkmark[/tex]

[tex]\huge\textsf{Good luck on your assignment \& enjoy your day!}[/tex]

~[tex]\textsf{Amphitrite1040:)}[/tex]

Answer:

Hence the answer is Letter B.

Step-by-step explanation:

° ° °

Answer:

This question seems to be asking for the work, so I put it below.

Anyhow, Marla had 3.5 candies, and Ron had 10.5.

Step-by-step explanation:

x = 3x - 7  (lets set this as "a")   { a = 3x - 7}

x + 7 = 3x -7 + 7

x + 7 = 3x

( x + 7 ) / 3 = 3/3x

( x + 7 ) / 3 = x     =    *(same as "a") {a}

( x + 7 ) / 3 = 3x - 7

( x + 7 ) / 3 * 3 = ( 3x - 7 ) * 3

x + 7 = 9x - 21

x - x + 7 = 9x - x - 21

7 = 8x - 21

7 + 21 = 8x - 21 + 21

28 = 8x

28 / 8 = 8/8x

3.5 = x

We can plug this back into the original equation and find that it is correct because:

x = 3x - 7

x = 3.5  =====

3.5 = 3( 3.5 ) - 7

3.5 = 10.5 - 7

3.5 = 3.5

Answer:

Marla has 7 and Ron has 21

Step-by-step explanation:

lets take

no. of candies Marla has as "x"

and no. of  candies Ron has will be "3x"

Marla says if Ron gives her seven candies, they will have the same no. of candies

so your equation will be -

x + 7 = 3x - 7 (as Marla gets the candy, Ron loses the candy)

3x - x = 7 + 7

2x = 14

∴x = 7

and 3x = 3 x 7 = 21

YOUR WELCOME

====================================================

Explanation:

The two smaller triangles are proportional, which lets us set up this equation

5/n = n/15

Cross multiplying leads to

5*15 = n*n

n^2 = 75

----------

Apply the pythagorean theorem on the smaller triangle on top, or on the right.

a^2+b^2 = c^2

5^2+n^2 = m^2

25+75 = m^2

100 = m^2

m^2 = 100

m = sqrt(100)

m = 10

7n-16

Step-by-step explanation:

Sry can u give me the picture

Answer:

2(2)+7=11

2(3)+7=13

2(4)+7=15

2(5)+7=17

2(9)+7=25

Step-by-step explanation:

2(2)+7

4+7=11

2(3)+7

6+7=13

2(4)+7

8+7=15

2(5)+7

10+7=17

2(9)+7

18+7=25

Step-by-step explanation:

Eueydhhdgdgdbdbddbdbhd

Answer:

Hello,

[tex]R=\dfrac{18}{\pi}\ (cm)[/tex]

Step-by-step explanation:

[tex]Formula: \ L=\theta*R\\[/tex]

[tex]R=\dfrac{6}{\dfrac{\pi}{3} } =\dfrac{6*3}{\pi} =\dfrac{18}{\pi}\ (cm)[/tex]

Answer:

In recreational mathematics, a square array of numbers, usually positive integers, is called a magic square if the sums of the numbers in each row, each column, and both main diagonals are the same.[1][2] The order of the magic square is the number of integers along one side (n), and the constant sum is called the magic constant. If the array includes just the positive integers {\displaystyle 1,2,...,n^{2}}{\displaystyle 1,2,...,n^{2}}, the magic square is said to be normal. Some authors take magic square to mean normal magic square.[3]

The smallest (and unique up to rotation and reflection) non-trivial case of a magic square, order 3

Magic squares that include repeated entries do not fall under this definition and are referred to as trivial. Some well-known examples, including the Sagrada Família magic square and the Parker square are trivial in this sense. When all the rows and columns but not both diagonals sum to the magic constant we have semimagic squares (sometimes called orthomagic squares).

The mathematical study of magic squares typically deals with its construction, classification, and enumeration. Although completely general methods for producing all the magic squares of all orders do not exist, historically three general techniques have been discovered: by bordering method, by making composite magic squares, and by adding two preliminary squares. There are also more specific strategies like the continuous enumeration method that reproduces specific patterns. Magic squares are generally classified according to their order n as: odd if n is odd, evenly even (also referred to as "doubly even") if n is a multiple of 4, oddly even (also known as "singly even") if n is any other even number. This classification is based on different techniques required to construct odd, evenly even, and oddly even squares. Beside this, depending on further properties, magic squares are also classified as associative magic squares, pandiagonal magic squares, most-perfect magic squares, and so on. More challengingly, attempts have also been made to classify all the magic squares of a given order as transformations of a smaller set of squares. Except for n ≤ 5, the enumeration of higher order magic squares is still an open challenge. The enumeration of most-perfect magic squares of any order was only accomplished in the late 20th century.

Magic squares have a long history, dating back to at least 190 BCE in China. At various times they have acquired occult or mythical significance, and have appeared as symbols in works of art. In modern times they have been generalized a number of ways, including using extra or different constraints, multiplying instead of adding cells, using alternate shapes or more than two dimensions, and replacing numbers with shapes and addition with geometric operations.

The sum area of shaded region​ is [tex]\frac{1}{3}[/tex].

The formula to find the sum of infinite geometric progression is

[tex]S_{n}=\frac{a_{1}(1-q^{n}) }{1-q} \ \ q \ne 1[/tex]

Given

S = [tex]\frac{1}{4} +\frac{1}{16} +\frac{1}{64} +.........[/tex]

Using geometric progression

S = [tex]\lim_{h \to \infty} [\frac{1}{4} +(\frac{1}{4} )^{2} +(\frac{1}{4} )^{3} +.........+(\frac{1}{4} )^{n}][/tex]

Using summation formula for geometric progression

[tex]S_{n}=\frac{a_{1}(1-q^{n}) }{1-q} \ \ q \ne 1[/tex]

= [tex]\lim_{h \to \infty} \frac{\frac{1}{4} (1-(\frac{1}{4} )^{n} }{1-\frac{1}{4} }[/tex]

= [tex]\lim_{h \to \infty} \frac{\frac{1}{4} (1-\frac{1}{4^{n} }) }{\frac{3}{4} }[/tex]

= [tex]\lim_{h\to \infty} \frac{1}{3}(1-\frac{1}{4^{n} } )[/tex]

[tex]\lim_{h\to \infty} \frac{1}{4^{n} }[/tex] = 0

S  = [tex]\frac{1}{3}(1-0) = \frac{1}{3}[/tex]

The sum area of shaded region​ is [tex]\frac{1}{3}[/tex].

Find out more information about summation formula for geometric progression here

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Answer:

y=16x

Step-by-step explanation:

24=1 1/2x=3/2x

x=24*(2/3)=16

y= total money earned; x=number of hours worked

Answer:

y-intercept is 0, x-intercept is 0 and 1

Step-by-step explanation:

For y-intercept, x=0 :

[tex]{ \tt{y = {(0)}^{2} - 2(0) }} \\ { \tt{y = 0}}[/tex]

For x-intercept, y=0 :

[tex]{ \tt{0 = {2x}^{2} - 2x }} \\ { \tt{2x(x - 1) = 0}} \\ { \tt{x = 0 \: \: and \: \: 1}}[/tex]