Please solve it <br><br> it's simultaneous equation involving logarithm

Please hurry i need help

kirza4 [7]

Answer:

(x^2+16x+63)

Step-by-step explanation:

(x^2+17x+72)-(1x+9)

(x^2+16x+63)

If this is incorrect I appologize

6 0

3 years ago

Read 2 more answers

Simplificar la radicacion 18

bonufazy [111]

Answer:

441

Step-by-step explanation:

8 0

1 year ago

Determine the slope-intercept form of the equation of the line parallel to y = -4/3 x + 11 that passes through the point (–6, 2)

MatroZZZ [7]

I've attached working :)

7 0

2 years ago

Find the following quotients, and write the quotient in standard form. (a)- x^2-9 ÷ x-3 (b)- x^3-27 ÷ x-3 (c) x^4- 81 ÷ x-3

IgorLugansk [536]

Answer:

(a) x+3

(b) $x^2+3x+9$

(c) $(x+3)(x^2+9)$

Step-by-step explanation:

We have given that

(a) $\frac{x^2-9}{x-3}$

From the algebraic identity we know that

$a^2-b^2=(a+b)(a-b)$

So $\frac{x^2-9}{x-3}=\frac{(x+3)(x-3)}{x-3}=x+3$

(b) $\frac{x^3-27}{x-3}$

We know the algebraic identity

$a^3-b^3=(a-b)(a^2+ab+b^2)$

So $\frac{x^3-27}{x-3}=\frac{x^3-3^3}{x-3}=\frac{(x-3)(x^2+3x+9)}{x-3}=x^2+3x+9$

(c) We have given $\frac{x^4-81}{x-3}$

We know the algebraic identity

$a^2-b^2=(a+b)(a-b)$

$\frac{x^4-81}{x-3}=\frac{(x^2)^2-(3^2)^2}{x-3}=\frac{(x^2-9)(x^2+9)}{x-3}=\frac{(x+3)(x-3)(x^2+9)}{x-3}=(x+3)(x^2+9)$

3 0

2 years ago

Show that x^2-y^2= (x+y)(x-y) with venn diagram

strojnjashka [21]

Answer:

$\sqrt[]{ \frac{0}{0} }$

Explainion are shown in pictures above

7 0

2 years ago

Please solve it it's simultaneous equation involving logarithm