Solve for x:<br> AD bisects ZBAC<br> A

I need help so much please help the attachment below is the question i need help on

leonid [27]

Answer:

$\dfrac{\sqrt[12]{55296}}{2}$

Step-by-step explanation:

Rationalize the denominator, then use a common root for the numerator.

$\dfrac{\sqrt[4]{6}}{\sqrt[3]{2}}=\dfrac{(2\cdot 3)^{\frac{1}{4}}}{2^{\frac{1}{3}}}\\\\=\dfrac{(2\cdot 3)^{\frac{1}{4}}}{2^{\frac{1}{3}}}\cdot\dfrac{2^{\frac{2}{3}}}{2^{\frac{2}{3}}}=\dfrac{2^{\frac{1}{4}+\frac{2}{3}}3^{\frac{1}{4}}}{2}\\\\=\dfrac{2^{\frac{11}{12}}3^{\frac{3}{12}}}{2}=\dfrac{\sqrt[12]{2^{11}3^{3}}}{2}\\\\=\dfrac{\sqrt[12]{55296}}{2}$

3 0

3 years ago

The fox population in a certain region has a continuous growth rate of 5% per year. It is estimated that the population in the y

kvasek [131]

Answer:

P(t) = A * (1 + r)^t ;

14,922 ;

Year 2013

Step-by-step explanation:

Given the following :

Continuous growth rate(r) = 5% = 0.05

Population in year 2000 = Initial population (A) = 10,100

Time(t) = period (years since year 2000)

A)

Find a function that models the population,P(t) , after (t) years since year 2000 (i.e. t= 0 for the year 2000).

P(t) = A * (1 + r)^t

Trying out our function for t = year 2000, t =0

P(0) = 10,100 * (1 + 0.05)^0

P(0) = 10,100 * 1.05^0 = 10,100

B.)

Use your function from part (a) to estimate the fox population in the year 2008.

Year 2008, t = 8

P(8) = 10,100 * (1 + 0.05)^8

P(8) = 10,100 * 1. 05^8

P(8) = 10,100 * 1.4774554437890625

= 14922.29

= 14,922

c) Use your function to estimate the year when the fox population will reach over 18,400 foxes. Round t to the nearest whole year, then state the year.

P(t) = A * (1 + r)^t

18400 = 10,100 * (1.05)^t

18400/10100 = 1.05^t

1.8217821 = 1.05^t

1.05^t = 1.8217821

In(1.05^t) = ln(1.8217821)

0.0487901 * t = 0.5998151

t = 0.5998151 / 0.0487901

t = 12.293787

Therefore eit will take 13 years

2000 + 13 = 2013

4 0