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2 years ago

What are the potential solutions of log4X+log4(x+6)=2?

Mathematics

2 answers:

aliya0001 [1]2 years ago

8 0

Answer:

Ox=-2 and x=-8

Step-by-step explanation:

Marat540 [252]2 years ago

5 0

Answer:

x = 2 ; x = -8

Step-by-step explanation:

$\sf \ log_4 \ x +log_4 \ (x +6) = 2\\\\ log_4 \ (x)*(x +6) = 2\\\\$

log₄ ( x² + 6x) = 2

4² = x² + 6x

x² + 6x - 16 = 0

x² - 2x + 8x - 16 = 0

x(x - 2) + 8(x - 2) = 0

(x - 2) (x + 8) = 0

x - 2 = 0 or x + 8 = 0

x = 2 or x = - 8

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What is the area of this triangle ?

oee [108]

Answer:

Area of triangle is 9.88 units^2

Step-by-step explanation:

We need to find the area of triangle

Given E(5,1), F(0,4), D(0,8)

We will use formula:

$Area\,\,of\,\,triangle =\sqrt{s(s-a)(s-b)s-c)} \\where\,\, s = \frac{a+b+c}{2}$

We need to find the lengths of side DE, EF and FD

Length of side DE = a = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Length of side DE = a = $=\sqrt{(5-0)^2+(1-8)^2}\\=\sqrt{(5)^2+(-7)^2}\\=\sqrt{25+49}\\=\sqrt{74}\\=8.60$

Length of side EF = b = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Length of side EF = b = $=\sqrt{(0-5)^2+(4-1)^2}\\=\sqrt{(-5)^2+(3)^2}\\=\sqrt{25+9}\\=\sqrt{34}\\=5.8$

Length of side FD = c = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Length of side FD = c = $=\sqrt{(0-0)^2+(8-4)^2}\\=\sqrt{(0)^2+(4)^2}\\=\sqrt{0+16}\\=\sqrt{16}\\=4$

so, a= 8.60, b= 5.8 and c = 4

s = a+b+c/2

s= 8.6+5.8+4/2

s= 9.2

Area of triangle= $=\sqrt{s(s-a)(s-b)s-c)}\\=\sqrt{9.2(9.2-8.6)(9.2-5.8)(9.2-4)}\\=\sqrt{9.2(0.6)(3.4)(5.2)}\\=\sqrt{97.5936}\\=9.88$

So, area of triangle is 9.88 units^2

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saveliy_v [14]

Answer:

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Step-by-step explanation:

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WINSTONCH [101]

Answer:

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