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2 years ago

What is the gcf of 60 and 72

1 answer:

4 0

       60             72
       /\                /\
     10x6           9x8
     /\     /\          /\    /\
2x5 2x3      3x3    2x4                  60 GCF= 2
                                    /\                  72 GCF = 2
                                    2x2

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What is the answer!?!?!?

77julia77 [94]

The price for hardcover books is 1.50 and the price for paperback is .50

8 0

3 years ago

Please prove this........

Crazy boy [7]

Answer: see proof below

Step-by-step explanation:

Given: A + B + C = π → C = π - (A + B)

→ sin C = sin(π - (A + B)) cos C = sin(π - (A + B))

→ sin C = sin (A + B) cos C = - cos(A + B)

Use the following Sum to Product Identity:

sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]

cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]

Use the following Double Angle Identity:

sin 2A = 2 sin A · cos A

Proof LHS → RHS

LHS: (sin 2A + sin 2B) + sin 2C

$\text{Sum to Product:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-\sin 2C$

$\text{Double Angle:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-2\sin C\cdot \cos C$

$\text{Simplify:}\qquad \qquad 2\sin (A + B)\cdot \cos (A - B)-2\sin C\cdot \cos C$

$\text{Given:}\qquad \qquad \quad 2\sin C\cdot \cos (A - B)+2\sin C\cdot \cos (A+B)$

$\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]$

$\text{Sum to Product:}\qquad 2\sin C\cdot 2\cos A\cdot \cos B$

$\text{Simplify:}\qquad \qquad 4\cos A\cdot \cos B \cdot \sin C$

LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C $\checkmark$

7 0

3 years ago

F(x) = (3x + 6) (x - 3)²

Sever21 [200]

f(x)= 3x³ - 18x +9

Algebraic identities are algebraic equations that are true regardless of the value of each variable. Additionally, they are employed in the factorization of polynomials. Algebraic identities are employed in this manner for the computation of algebraic expressions and the solution of various polynomials.

Identity I: (a + b)² = a² + 2ab + b²

Identity II: (a – b)² = a² – 2ab + b²

Identity III: a² – b²= (a + b)(a – b)

Identity IV: (x + a)(x + b) = x² + (a + b) x + ab

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

Identity VI: (a + b)³ = a³ + b³ + 3ab (a + b)

Identity VII: (a – b)³ = a³ – b³ – 3ab (a – b)

Identity VIII: a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)

f(x) = (3x + 6) (x - 3)²

= ( 3x + 6) ( x - 3 )²

= ( 3x + 6)( x² - 6x + 9)

= 3x( x² - 6x + 9) + 6( x² - 6x + 9)

= 3x³ - 6x² + 18x + 6x² - 36x +9

= 3x³ - 18x +9

To learn more about algebraic expansions, refer to brainly.com/question/4344214

#SPJ9

5 0

2 years ago

Question 2 Please help. Find the arc length for the following:

harkovskaia [24]

9.4 would be your answer

4 0

2 years ago

Read 2 more answers

When solving the equation x2 – 8x - 7 = 0 by completing the square, which equation

IgorLugansk [536]

<h3>Answer:-</h3>

Lets Solve

$\\ \rm \longmapsto \: {x}^{2} - 8x - 7 = 0 \\ \\ \rm \longmapsto \: {x}^{2} - 8 {}^{}x = 7 \\ \\ \rm \longmapsto \: 4( {x}^{2} - 8x) = 4 \times 7 \\ \\ \rm \longmapsto \: {4x}^{2} - 32x = 28 \\ \\ \rm \longmapsto \: {(2x)}^{2} - 2 \times 2x \times 8 = 28 \\ \\ \rm \longmapsto \: (2x) {}^{2} - 2 \times 2x \times 8 + {8}^{2} = 28 + {8}^{2} \\ \\ \rm \longmapsto \: {(2x - 8)}^{2} = 64 + 28 \\ \\ \rm \longmapsto \: {(2x - 8)}^{2} = 92 \\ \\ \rm \longmapsto \: 2x - 8 = \underline{ + }9.3 \\ \\ \rm \longmapsto \:2x = 8 + 9.3 \: or \: 8 - 9.3 \\ \\ \rm \longmapsto \:2x = 17.3 \:or \: - 1.3 \\ \\ \rm \longmapsto \:2x = 17.3 \\ \\ \rm \longmapsto \:x = \frac{17.3}{2} \\ \\ \rm \longmapsto x = \:8.6$

Ignore negative value

4 0

3 years ago