All categories

2 years ago

Which of the following segments is an angle bisector? 1. AL 2. JB 3. KC

Mathematics

1 answer:

Orlov [11]2 years ago

4 0

Answer:

I think it's 3) KC

Hopefully it's correct

You might be interested in

Solve ln 2 + ln x = 5. Round to the nearest thousandth, if possible.

azamat

Answer:

The value of x nearest to thousandths is, 74.207

Step-by-step explanation:

Solve : $\ln 2 + \ln x = 5$

$\ln$ represents the logarithmic function with base e.

Using logarithmic properties:

$\ln (mn) = \ln m + \ln n$

$\ln x = a$ , then $x = e^a$

$\ln 2 + \ln x = 5$

Apply the logarithmic properties; we have

$\ln 2x = 5$

then;

$2x = e^5$

$2x=148.41315910258$

Divide both sides by 2 we get;

$x = 74.2065796$

Therefore, the value of x nearest to thousandths is, 74.207

3 0

3 years ago

How do I write 125% in simplest form

Lera25 [3.4K]

Answer:

The answer is 5/4 in fraction form and 1.25 decimal form

7 0

2 years ago

Brainliest!!

AlladinOne [14]

Answer:

Step-by-step explanation:

BC supplementary means they add up to 180 and answer d adds up to 180

6 0

2 years ago

Please see attachment

Dafna11 [192]

Answer:

a) The value of absolute minimum value = - 0.3536

b) which is attained at $x = \frac{1}{\sqrt{2} }$

Step-by-step explanation:

Step(i):-

Given function

$f(x) = \frac{-x}{2x^{2} +1}$ ...(i)

Differentiating equation (i) with respective to 'x'

$f^{l} = \frac{2x^{2} +1(-1) - (-x) (4x)}{(2x^{2}+1)^{2} }$ ...(ii)

$f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} }$

Equating Zero

$f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

$\frac{2x^{2}-1}{(2x^{2}+1)^{2} } = 0$

$2 x^{2}-1 = 0$

$2 x^{2} = 1$

$x^{2} = \frac{1}{2}$

$x = \frac{-1}{\sqrt{2} } , x = \frac{1}{\sqrt{2} }$

Step(ii):-

Again Differentiating equation (ii) with respective to 'x'

$f^{ll}(x) = \frac{(2x^{2} +1)^{2} (4x) - 2(2x^{2} +1) (4x)(2x^{2}-1) }{(2x^{2}+1)^{4} }$

put

$x = \frac{1}{\sqrt{2} }$

$f^{ll} (x) > 0$

The absolute minimum value at $x = \frac{1}{\sqrt{2} }$

Step(iii):-

The value of absolute minimum value

$f(x) = \frac{-x}{2x^{2} +1}$

$f(\frac{1}{\sqrt{2} } ) = \frac{-\frac{1}{\sqrt{2} } }{2(\frac{1}{\sqrt{2} } )^{2} +1}$

on calculation we get

The value of absolute minimum value = - 0.3536

Final answer:-

a) The value of absolute minimum value = - 0.3536

b) which is attained at $x = \frac{1}{\sqrt{2} }$

3 0

2 years ago

18.

chubhunter [2.5K]

A bakers dozen is 12 which would be $0.41 for each cookie. If you meant 13, it would be $0.38. All you have to do is divide 4.94 by the amount of cookies

7 0

2 years ago