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Thepotemich [5.8K]
2 years ago
9

Does anyone know when the tutors is coming for Brainly ?

Mathematics
1 answer:
frosja888 [35]2 years ago
7 0

Answer:

I have no idea it does show anything about it I checked just rely on the students to help you until they show up :)

You might be interested in
In​ 2012, the population of a city was 6.27 million. The exponential growth rate was 1.47​% per year.
Tcecarenko [31]

Answer:

a) a(t)= 6.27e^{0.0147*t}

b) 6.84 million

c) 2028.52

d) 47 years

Step-by-step explanation:

a)Here we know standard exponential growth rate is

a(t)= ae^{k*t}

In this put t=0 and a(t)=6.27 million

thus we get a = 6.27 million

now \frac{da}{dt}=a(t)k

  this is equal to(in 2012)=\frac{1.47}{100}*=0.0147

b)Put t = 6

we get

a(t)= 6.27e^{0.0147*6}

=6.84 million

c) 8 = 6.27e^{0.0147*t}

  take log both sides

we get 16.52 years

d) 2*6.27=6.27e^{0.0147*t}

take log both sides we get 47 years to get double the current population.

3 0
3 years ago
Consider the function represented by the tabe what is f(0)
lisov135 [29]
The correct answer above
3 0
3 years ago
−8.5x + 0.36 = −3.04
erma4kov [3.2K]
X= 0.4

-8.5x=-3.4
x=0.4
4 0
3 years ago
A total of 691 tickets were sold for the school play they were either adult tickets or student tickets there were fifty nine few
garri49 [273]
691 = A+ S
A=S+59
691 = S+59+S
691=2S+59
691-59 = 2S +59 -59
632 = 2S
316= S
5 0
3 years ago
2. Consider the circle x^2−4x+y^2+10y+13=0. There are two lines tangent to this circle having a slope of 2/3.
likoan [24]

Answer:

a) The coordinates are (0.431, -2.646) and (3.568,-7.354)

b) The tangent lines are

l₁ = (x-0.431)*2/3 - 2.646

l₂ = (x-3.568)2/3 - 7.354

Step-by-step explanation:

First, lets complete squares, by taking for each cordinate the square of a linear expression

0= x²−4x+y²+10y+13 = (x-2)²+ 4  + (y+5)²-25+13 = (x-2)²+(y+5)² - 8

Hence (x-2)² + (y+5)² = 8

Lets put y in function of x. We should obtain 2 functions f and g that represent the circle.

(y+5)² = 8 - (x-2)² = -x² + 4x + 4

y = ^+_- \sqrt{-x^2+4x+4} -5

Thus

f(x) = \sqrt{-x^2+4x+4} - 5

g(x) = -\sqrt{-x^2+4x+4} - 5

Lets find the derivate of each function and the points in which they reach the value 2/3. In those points the tangent line will have a slope of 2/3.

We may just find the values of x which the derivate of f is either 2/3 or -2/3.

\frac{-x+2}{\sqrt{-x^2+4x+4}} = ^+_-\frac{2}{3} \Leftrightarrow \frac{x^2-4x+4}{-x^2+4x+4} = \frac{4}{9} \Leftrightarrow 9(x^2-4x+4) = 4(-x^2+4x+4) \\\Leftrightarrow 13x^2-52x+20 = 0

The quadratic has roots

\frac{52 ^+_-\sqrt{1664}}{26}

one root is 3.568, which corresponds with g, and the other root is 0.431, corresponding to f.

Also g(3.568) = - 7.354 and f(0.431) = -2.646

This means that the coordinates are (0.431 , -2.646), (3.568 , -7.354) and the tangent lines are

l₁ = (x-0.431)*2/3 - 2.646

l₂ = (x-3.568)2/3 - 7.354

5 0
3 years ago
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