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4 years ago

5

Can you help me please

1 answer:

STALIN [3.7K]4 years ago

4 0

Answer:

36

Step-by-step explanation:

Here we a given a rule

It say

$x(y$ ÷ $3)^2$

And we are given the values of x and y as

x=4 and y = 9

Let us replace them in the rule we are provided with

$4(9$ ÷ $3)^2$

Applying DMAS rule, which says, first Division ollowed by Multiplication addition and then subtraction .

We first divide 9 by 3

9÷3 = 3

Hence

$4(9$ ÷ $3)^2 = 4(3)^2$

Now we square first

$3^2 = 3 \times 3 = 9$

$4(3)^2 = 4 \times 9$

$4(3)^2 = 36$

Hence our final answer will be 36

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Try this hard Math Problem if you dare!!

Dennis_Churaev [7]

Answer:

a. $(x - 3)^2 + 16$

b. $8(x -7)^2$

c. $(a^2 - 1)(7x - 6)$ or $(a+1)(a-1)(7x-6)$

d. $(x^2-4)(x^2+3)$ or $(x-2)(x+2)(x^2+3)$

e. $(a^n+b^n)(a^n-b^n)(a^{2n} +b^{2n})$

Step-by-step explanation:

$a.\ (x + 1)^2 - 8(x - 1) + 16$

Expand

$(x + 1)(x + 1) - 8(x - 1) + 16$

Open brackets

$x^2 + x + x + 1 - 8x + 8 + 16$

$x^2 + 2x + 1 - 8x + 24$

Collect Like Terms

$x^2 + 2x - 8x+ 1 + 24$

$x^2 - 6x+ 25$

Express 25 as 9 + 16

$x^2 - 6x+ 9 + 16$

Factorize:

$x^2 - 3x - 3x + 9 + 16$

$x(x -3)-3(x - 3) + 16$

$(x - 3)(x - 3) + 16$

$(x - 3)^2 + 16$

$b.\ 8(x - 3)^2 - 64(x-3) + 128$

Expand

$8(x - 3)(x - 3) - 64(x-3) + 128$

$8(x^2 - 6x+ 9) - 64(x-3) + 128$

Open Brackets

$8x^2 - 48x+ 72 - 64x+192 + 128$

Collect Like Terms

$8x^2 - 48x - 64x+192 + 128+ 72$

$8x^2 -112x+392$

Factorize

$8(x^2 -14x+49)$

Expand the expression in bracket

$8(x^2 -7x-7x+49)$

Factorize:

$8(x(x -7)-7(x-7))$

$8((x -7)(x-7))$

$8(x -7)^2$

$c.\ 7a^2x - 6a^2 - 7x + 6$

Factorize

$a^2(7x - 6) -1( 7x - 6)$

$(a^2 - 1)(7x - 6)$

The answer can be in this form of further expanded as follows:

$(a^2 - 1^2)(7x - 6)$

Apply difference of two squares

$(a+1)(a-1)(7x-6)$

$d.\ x^4 - x^2 - 12$

Express $x^4$ as $x^2$

$(x^2)^2 - x^2 - 12$

Expand

$(x^2)^2 +3x^2- 4x^2 - 12$

$x^2(x^2+3) -4(x^2+3)$

$(x^2-4)(x^2+3)$

The answer can be in this form of further expanded as follows:

$(x^2-2^2)(x^2+3)$

Apply difference of two squares

$(x-2)(x+2)(x^2+3)$

$e.\ a^{4n} -b^{4n}$

Represent as squares

$(a^{2n})^2 -(b^{2n})^2$

Apply difference of two squares

$(a^{2n} -b^{2n})(a^{2n} +b^{2n})$

Represent as squares

$((a^{n})^2 -(b^{n})^2)(a^{2n} +b^{2n})$

Apply difference of two squares

$(a^n+b^n)(a^n-b^n)(a^{2n} +b^{2n})$

6 0

3 years ago

(-10x)(2) is there any chance you can help please and thanks

Svetllana [295]

The answer is -20x, because the parentheses show they are being multiplied, so you multiply -10x and 2, which would be -20x.

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3 years ago

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Answer:

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Step-by-step explanation:

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Answer:

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