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3 years ago

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Find the distance between A and B. A(12,-1) B(0,-6)

1 answer:

Sonbull [250]3 years ago

4 0

$The\ distance\ between\ A(x_A;\ y_A)\ and\ B(x_B;\ y_B):\\\\d=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}\\\\------------------------------\\\\A(12;-1);\ B(0;-6)\\\\subtitute\\\\d=\sqrt{(0-12)^2+(-6-(-1))^2}=\sqrt{(-12)^2+(-5)^2}\\\\=\sqrt{144+25}=\sqrt{169}=\boxed{13}$

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Find the interval(s) where the function is increasing and the interval(s) where it is decreasing. (Enter your answers using inte

Mashcka [7]

Answer:

f'(x) > 0 on $(-\infty ,\frac{1}{e})$ and f'(x)<0 on $(\frac{1}{e},\infty)$

Step-by-step explanation:

1) To find and interval where any given function is increasing, the first derivative of its function must be greater than zero:

$f'(x)>0$

To find its decreasing interval :

$f'(x)$

2) Then let's find the critical point of this function:

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[6-2^{2x}]=\frac{\mathrm{d} }{\mathrm{d}x}[6]-\frac{\mathrm{d}}{\mathrm{d}x}[2^{2x}]=0-[ln(2)*2^{2x}*\frac{\mathrm{d}}{\mathrm{d}x}[2x]=-ln(2)*2^{2x}*2=-ln2*2^{2x+1\Rightarrow }f'(x)=-ln(2)*2^{2x}*2\\-ln(2)*2^{2x+1}=-2x^{2x}(ln(x)+1)=0$

2.2 Solving for x this equation, this will lead us to one critical point since x' is not defined for Real set, and x'' $=\frac{1}{e}$ ≈0.37 for e≈2.72

$x^{2x}=0 \ni \mathbb{R}\\(ln(x)+1)=0\Rightarrow ln(x)=-1\Rightarrow x=e^{-1}\Rightarrow x\approx 2.72^{-1}x\approx 0.37$

3) Finally, check it out the critical point, i.e. f'(x) >0 and below f'(x)<0.

8 0

3 years ago

570 divided by 94 what is the answer

Tanzania [10]

6.063... If you use a calculator you'll see that the number doesn't end/repeat.

3 0

3 years ago

Read 2 more answers

Daniel had $57 to spend on gifts he spent $10.51 on a gift for Isaac $8.73 on a gift for madison and $8.47 on a gift for Abigail

mina [271]

57-10.51=46.49
46.49-8.73=37.76
37.76-8.47=29.29
Therefore Daniel has $29.29 left.

7 0

3 years ago

Read 2 more answers

a random sample of high school seniors were asked whether they were applying for college. the resulting confidence interval for

Artist 52 [7]

Using the confidence interval, the margin of error is 0.02.

In the given equation we have to find the margin of error.

The resulting confidence interval for the proportion of students applying for college is (0.65,0.69).

Let p = population proportion of students applying for college

P = Sample proportion of students applying for college

ME = margin of error

So the confidence interval

CI = (P-ME, P+ME)

From the given question;

(P-ME, P+ME) = (0.65,0.69)

So lower bound= P-ME = 0.65............................(1)

Upper Bound = P+ME = 0.69............................(2)

Simplifying the Equation 1 and 2

Add equation 1 and 2 we get

2P = 1.34

Divide by 2 on both side we get

P = 0.67

Putting the value of P in the equation 2

P+ME = 0.69

0.67+ME=0.69

Subtract 0.67 on both side we get

ME = 0.02

Hence, the margin of error is 0.02.

To learn more about margin of error link is here

brainly.com/question/10501147

#SPJ4

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1 year ago

I need help ASAP please

Anna007 [38]

I can’t see the question

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3 years ago