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3 years ago

Which function described below has the greatest rate of change? I WILL MARK BRAINLIEST

Mathematics

1 answer:

Ymorist [56]3 years ago

8 0

Answer:

C III

Step-by-step explanation:

The rate of change of a linear function is the slope.

f(x) = mx + b is the equation of a linear function whose graph is a straight line. m is the slope.

I f(x) = 4x - 3; m = slope = 4

II f(x) = 1/2 x + 5; m = slope = 1/2

III We can use two points to find the slope.

Let's use points (1, 6) and (2, 12).

m = slope = (y2 - y1)/(x2 x1) = (12 - 6)/(2 - 1) = 6/1 = 6

The three slopes are 4, 1/2, 6.

The greatest rate of change is 6, so the answer is C III.

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Anybody good in geometry and know how to do this? Free Brainliest and points !! I need at least 3 ppl to answer this !

yan [13]

Answer:

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Step-by-step explanation:

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3 years ago

Express sin A,cos A and tan A as ratios

OverLord2011 [107]

Answer:

Part A) $sin(A)=\frac{2\sqrt{42}}{23}$

Part B) $cos(A)=\frac{19}{23}$

Part C) $tan(A)=\frac{2\sqrt{42}}{19}$

Step-by-step explanation:

Part A) we know that

In the right triangle ABC of the figure the sine of angle A is equal to divide the opposite side angle A by the hypotenuse

$sin(A)=\frac{BC}{AB}$

substitute the values

$sin(A)=\frac{2\sqrt{42}}{23}$

Part B) we know that

In the right triangle ABC of the figure the cosine of angle A is equal to divide the adjacent side angle A by the hypotenuse

$cos(A)=\frac{AC}{AB}$

substitute the values

$cos(A)=\frac{19}{23}$

Part C) we know that

In the right triangle ABC of the figure the tangent of angle A is equal to divide the opposite side angle A by the adjacent side angle A

$tan(A)=\frac{BC}{AC}$

substitute the values

$tan(A)=\frac{2\sqrt{42}}{19}$

8 0

3 years ago

Find the circumfrance and the area of a circle with a 5 ft radius. Use the value 3.14

Crazy boy [7]

Answer:

Circumference is 2piR =31.4

Area is pir^2=78.5

Step-by-step explanation:

6 0

2 years ago

In quadratic drag problem, the deceleration is proportional to the square of velocity

Mars2501 [29]

Part A

Given that $a= \frac{dv}{dt} =-kv^2$

Then,

$\int dv= -kv^2\int dt \\ \\ \Rightarrow v(t)=-kv^2t+c$

For $v(0)=v_0$ , then

$v(0)=-kv^2(0)+c=v_0 \\ \\ \Rightarrow c=v_0$

Thus, $v(t)=-kv(t)^2t+v_0$

For $v(t)= \frac{1}{2} v_0$ , we have

$\frac{1}{2} v_0=-k\left( \frac{1}{2} v_0\right)^2t+v_0 \\ \\ \Rightarrow \frac{1}{4} kv_0^2t=v_0- \frac{1}{2} v_0= \frac{1}{2} v_0 \\ \\ \Rightarrow kv_0t=2 \\ \\ \Rightarrow t= \frac{2}{kv_0}$

Part B

Recall that from part A,

$v(t)= \frac{dx}{dt} =-kv^2t+v_0 \\ \\ \Rightarrow dx=-kv^2tdt+v_0dt \\ \\ \Rightarrow\int dx=-kv^2\int tdt+v_0\int dt+a \\ \\ \Rightarrow x=- \frac{1}{2} kv^2t^2+v_0t+a$

Now, at initial position, t = 0 and $v=v_0$ , thus we have

$x=a$

and when the velocity drops to half its value, $v= \frac{1}{2} v_0$ and $t= \frac{2}{kv_0}$

Thus,

$x=- \frac{1}{2} k\left( \frac{1}{2} v_0\right)^2\left( \frac{2}{kv_0} \right)^2+v_0\left( \frac{2}{kv_0} \right)+a \\ \\ =- \frac{1}{2k} + \frac{2}{k} +a$

Thus, the distance the particle moved from its initial position to when its velocity drops to half its initial value is given by

$- \frac{1}{2k} + \frac{2}{k} +a-a \\ \\ = \frac{2}{k} - \frac{1}{2k} = \frac{3}{2k}$

7 0

3 years ago

The population of a town 4 years ago was 50,000

tino4ka555 [31]

Answer:

population will be 79806.

Step-by-step explanation:

Population of a town 4 years ago was 50,000

Population of town had grown exponentially and present population is 63,124.

So the formula of the population growth will be

$A_{n}=A_{o}(r)^{n-1}$