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3 years ago

Porfa ayuda :c t.t u.u

Mathematics

1 answer:

Orlov [11]3 years ago

4 0

Answer: 40

ADE = 70 por suma de angulos interiores en triangulo ADE

En triangulos BDE y AED

AE=BE por dato

DE lado comun

AED= BED= 90

por tanto los triangulos son iguales por tener dos lados y el angulo comprendido respectivamente iguales

Luego ADE = BDE = 70 y AD=BD=BC por ser triangulos BDE y AED iguales

Luego triangulo BDC isosceles de base DC

ADE + BDE + BDC = 180 por suma de angulos consecutivos

BDC = 180 - 70 -70 = 40

Ф = BDC =BCD = 40 por ser angulos bases de un triangulo isosceles

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Answer:

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Step-by-step explanation:

3x^3+3x^2-4x-4=0

By factoring, I can get simpler equations to get the roots.

x + 1 = 0, or x = -1

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3 years ago

After 3000 impressions, ad A has a CTR of 2% and a conversion rate of 25%, and ad B has a CTR of 4% and a conversion rate of 10%

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3 years ago

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3 years ago

What are the potential solutions of log4x+log4(x+6)=2?

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The potential solutions of $log_4x+log_4(x+6)=2$ are 2 and -8.

<h3>Properties of Logarithms</h3>

From the properties of logarithms, you can rewrite logarithmic expressions.

The main properties are:

Product Rule for Logarithms - $log_{b}(a*c)=log_{b}a+log_{b}c$
Quotient Rule for Logarithms - $log_{b}(\frac{a}{c} )=log_{b}a-log_{b}c$
Power Rule for Logarithms - $log_{b}(a^c)=c*log_{b}a$

The exercise asks the potential solutions for $log_4x+log_4(x+6)=2$ . In this expression you can apply the Product Rule for Logarithms.

$log_4x+log_4(x+6)=2\\ \\ x*(x+6)=4^2\\ \\ x^2+6x=16\\ \\ x^2+6x-16=0$

Now you should solve the quadratic equation.

Δ= $b^2-4ac=36-4*1*(-16)=36+64=100$ . Thus, x will be $x_{1,\:2}=\frac{-6\pm \:\sqrt{100} }{2\cdot \:1}=\frac{-6\pm \:10}{2}$ . Then:

$x_1=\frac{-6+10}{2}=\frac{4}{2} =2\\ \\ \:x_2=\frac{-6-10}{2}=\frac{-16}{2} =-8$

The potential solutions are 2 and -8.

Read more about the properties of logarithms here:

brainly.com/question/14868849

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