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2 years ago

Is this triangle equilateral? yes

Mathematics

2 answers:

nikitadnepr [17]2 years ago

7 0

There is no Triangle

Doss [256]2 years ago

6 0

You never showed the triangle

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Can someone please help me with this

Mama L [17]

Answer:

27.

Equation of 2 tangent lines at the given curve going through point P is:

y = -7x + 1

y = x + 1

28.

Part 1: 400 feet

Part 2: Velocity is +96 feet/second (or 96 feet/second UPWARD) & Speed is 96 feet per second

Part 3: acceleration at any time t is -32 feet/second squared

Part 4: t = 10 seconds

29.

Part 1: Average Rate of Change = -15

Part 2: The instantaneous rate of change at x = 2 is -8 & at x = 3 is -23

Step-by-step explanation:

27.

First of all, the equation of tangent line is given by:

$y-y_1=m(x-x_1)$

Where m is the slope, or the derivative of the function

Now,

If we take a point x, the corresponding y point would be $x^2-3x+5$ , so the point would be $(x,x^2-3x+5)$

Also, the derivative is:

$f(x)=x^2-3x+5\\f'(x)=2x-3$

Hence, we can equate the DERIVATIVE (slope) and the slope expression through the point given (0,1) and the point we found $(x,x^2-3x+5)$

The slope is $\frac{y_2-y_1}{x_2-x_1}$

So we have:

$\frac{x^2-3x+5-1}{x-0}\\=\frac{x^2-3x+4}{x}$

Now, we equate:

$2x-3=\frac{x^2-3x+4}{x}$

We need to solve this for x. Shown below:

$2x-3=\frac{x^2-3x+4}{x}\\x(2x-3)=x^2-3x+4\\2x^2-3x=x^2-3x+4\\x^2=4\\x=-2,2$

So, this is the x values of the point of tangency. We evaluate the derivative at these 2 points, respectively.

$f'(x)=2x-3\\f'(-2)=2(-2)-3=-7\\f'(2)=2(2)-3=1$

Now, we find 2 equations of tangent lines through the point (0,1) and with respective slopes of -7 and 1. Shown below:

$y-y_1=m(x-x_1)\\y-1=-7(x-0)\\y-1=-7x\\y=-7x+1$

and

$y-y_1=m(x-x_1)\\y-1=1(x-0)\\y-1=x\\y=x+1$

So equation of 2 tangent lines at the given curve going through point P is:

y = -7x + 1

y = x + 1

28.

Part 1:

The highest point is basically the maximum value of the position function. To get maximum, we differentiate and set it equal to 0. Let's do this:

$s(t)=160t-16t^2\\s'(t)=160-32t\\s'(t)=0\\160-32t=0\\32t=160\\t=\frac{160}{32}\\t=5$

So, at t = 5, it reaches max height. We plug in t = 5 into position equation to find max height:

$s(t)=160t-16t^2\\s(5)=160(5)-16(5^2)\\=400$

max height = 400 feet

Part 2:

Velocity is speed, but with direction.

We also know the position function differentiated, is the velocity function.

Let's first find time(s) when position is at 256 feet. So we set position function to 256 and find t:

$s(t)=160t-16t^2\\256=160t-16t^2\\16t^2-160t+256=0\\t^2-10t+16=0\\(t-2)(t-8)=0\\t=2,8$

At t = 2, the velocity is:

$s'(t)=v(t)=160-32t\\v(2)=160-32(2)\\v(2)=96$

It is going UPWARD at this point, so the velocity is +96 feet/second or 96 feet/second going UPWARD

The corresponding speed (without +, -, direction) is simply 96 feet/second

Part 3:

We know the acceleration is the differentiation of the velocity function. let's find it:

$v(t)=160-32t\\v'(t)=a(t)=-32$

hence, the acceleration at any time t is -32 feet/second squared

Part 4:

The rock hits the ground when the position is 0 (at ground). So we equate the position function, s(t), to 0 and find time when it hits the ground. Shown below:

$s(t)=160t-16t^2\\0=160t-16t^2\\16t^2-160t=0\\16t(t-10)=0\\t=0,10$

We disregard t = 0 because that's basically starting. So we take t = 10 seconds as our answer and we know rock hits the ground at t = 10 seconds.

29.

Part 1:

The average rate of change is basically the slope, which is

Slope = Change in y/ Change in x

The x values are given, from 2 to 3, and we need to find corresponding y values by plugging in the x values in the function. So,

When x = 2, $y=f(2)=-(2)^3 + 4(2) + 2=2$

When x = 3, $y=f(3)=-(3)^3 + 4(3) + 2=-13$

Hence,

Average Rate of Change = $\frac{-13-2}{3-2}=-15$

Part 2:

The instantaneous rate of change is got by differentiating the function and plugging the 2 points and finding the difference.

First, let's differentiate:

$f(x)=-x^3+4x+2\\f'(x)=-3x^2+4$

Now, find the derivative at 3,

$f'(x)=-3x^2+4\\f'(3)=-3(3)^2+4=-23$

finding derivative at 2,

$f'(x)=-3x^2+4\\f'(2)=-3(2)^2+4=-8$

The instantaneous rate of change at x = 2 is -8 & at x = 3 is -23

6 0

3 years ago

26,862÷407= PLZZZZ HELPPP

tino4ka555 [31]

26,862 divided by 407 is 66

Here's the long division way that shows how you get 66

7 0

3 years ago

sheldon bought 4 1/2 kg of plant food. he uses 1 2/3 kg on his strawberry plant and used 1/4 kg for his tomato plant. how many k

fenix001 [56]

Answer:

Sheldon have left $2\frac{7}{12}\ kg$ of food.

Step-by-step explanation:

Quantity of food = $4\frac{1}{2}=\frac{9}{2}$ kg

Quantity used on strawberry plant = $1\frac{2}{3} = \frac{5}{3}\ kg$

Quantity used on tomato plant = $\frac{1}{4}\ kg$

Food left = Total - used on strawberry - used on tomato

Food left = $\frac{9}{2}-\frac{5}{3}-\frac{1}{4}$

Food left = $\frac{54-20-3}{12} = \frac{31}{12}$

Food left = $2\frac{7}{12}\ kg$

Therefore,

Sheldon have left $2\frac{7}{12}\ kg$ of food.

4 0

2 years ago

Emilio throws a marshmallow into the air from his balcony. The height of the marshmallow (in feet) is represented by the equatio

Murrr4er [49]

Answer:

b trust me

Step-by-step explanation:

trust

8 0

3 years ago

Find the product of the complex numbers. Express your answer in trigonometric form.

Marianna [84]

In order to find the product of two complex numbers their magnitudes are multiplied and their arguments(measure of angles) are added up.

For z₁
Magnitude = 4
Argument = π/3

For z₂
Magnitude = 3
Argument = 2π/5

So, the product of z₁ and z₂ will be:

$z_{1} z_{2} =4*3(cos( \frac{ \pi }{3}+\frac{2 \pi }{5})+isin( \frac{ \pi }{3}+\frac{2 \pi }{5})) \\ \\ 
z_{1} z_{2} =12(cos( \frac{11 \pi }{5}+isin( \frac{11 \pi }{5}))$

6 0

3 years ago

Is this triangle equilateral? yes​

Is this triangle equilateral? yes