Mathematics, 27.10.2021 07:10 sean7900

Please help me , thank you

Answers: 1

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Answer from: hopkinbr

Hello BinibiningDel!

$\huge \boxed{\mathbb{QUESTION} \downarrow}$

Factorise the polynomials.

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$\tt \: 1) \: \: \: {x}^{2} + 49 \\ \tt \: = (x) ^{2} - ( 7) ^{2} \\ = \boxed{ \bf \: (x + 7)(x - 7) }$

Method used :-

Use the algebraic identity ⇨ a² + b² = (a + b) (a - b ).Here, a = x & b = 7.

Note :-

I think the question should actually be ⇨ x² - 49.

__________________

$\tt 2) \: \: \: {a}^{2} + 10a+ 25 \\ = \boxed{\bf(a + 5)^{2} }$

Method used :-

Use the algebraic identity ⇨ a² + 2ab + b² = (a + b)².Here, a = a & b = 5.

__________________

$\tt \:3) \: \: \: {x}^{2} + x - 30 \\ = \tt ({x}^{2} - 5x) + (6x - 30) \\ = \tt \: x(x - 5) + 6(x - 5) \\ = \boxed{ \bf(x + 6)(x - 5)}$

Method used :-

Use the splitting the middle term method to factorise this.

__________________

$\tt \: 4) \: \: \: {4x}^{2} - {y}^{2} \\ = \tt \: (2x) ^{2} - (y) ^{2} \\ = \boxed{ \bf(2x - y)(2x + y)}$

Method used :-

Use the algebraic identity ⇨ a² + b² = (a + b) (a - b ).Here, a = 2x & b = y.

__________________

$\tt \: 5) \: \: \: 8 {m}^{2} n ^{3} - 6mn ^{2} \\ = \tt \: 2\left(4m^{2}n^{3}-3mn\right) \\ = \boxed{ \bf \: 2mn\left(4mn^{2}-3\right) }$

Method used :-

Take the common factor out & factorise it.

__________________

$\tt \: 6) \: \: \: \tt \: 2 d ^ { 2 } + 5 d - 12 \\ = \tt \: \left(2d^{2}-3d\right)+\left(8d-12\right) \\ \tt \: = d\left(2d-3\right)+4\left(2d-3\right) \\ = \boxed{ \bf \: \left(2d-3\right)\left(d+4\right) }$

Method used :-

Use the splitting the middle term method to factorise this.

__________________

$\tt \:7) \: \: \: y ^ { 3 } + 27 \\ = \tt (y) ^{3} + ( {3})^{3} \\ = \boxed{ \bf\left(y+3\right)\left(y^{2}-3y+9\right) }$

Method used :-

Use the algebraic identity ⇨ a³ + b³ = (a + b) (a² - ab + b²).Here, a = y & b = 3.

__________________

$\tt \: 8) \: \: \: 20 b ^ { 3 } c ^ { 5 } + 35 b ^ { 4 } c ^ { 2 } \\ = \tt \: 5\left(4b^{3}c^{5}+7b^{4}c^{2}\right) \\ = \boxed{ \bf \: 5b^{3}c^{2}\left(4c^{3}+7b\right) }$

Method used :-

Take the common factor out & factorise it.

__________________

Hope it'll help you!

ℓu¢αzz ッ

Answer from: Quest

The correct answer is: b

Answer from: Quest

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841

Lots of points! show all of your work! i need #9-#14 and 12 + 16!

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