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Nina [5.8K]
2 years ago
15

Giving you 50 points if correct!!

Mathematics
1 answer:
egoroff_w [7]2 years ago
7 0
The answer is 472.441
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Solve the following using Substitution Method. <br> c= 3d - 27<br> 4d + 10c =120
svp [43]

Answer:

d=30-5c/2

Step-by-step explanation:

Move all terms that don't contain d to the right side and solve

8 0
3 years ago
Is percent of change less than 100
Ivanshal [37]
You can't change more than 100 percent
6 0
2 years ago
Applying the Third Corollary to the Inscribed Angles Theorem A circle is shown. Points A, B, C, and D are on the circle. Lines c
Marat540 [252]

Answer:

  • ∠B = 110°
  • ∠D = 70°

Step-by-step explanation:

Opposite angles in an inscribed quadrilateral are supplementary, so ...

  ∠B +∠D = 180°

  (3x -4) +(2x -6) = 180

  5x = 190 . . . . . . . . . . . add 10, collect terms

  x = 38 . . . . . . . . . . . . . divide by 5

  ∠D = 2(38) -6 = 70

  ∠B = 180 -70 = 110

The measures of angles B and D are 110° and 70°, respectively.

8 0
3 years ago
Can someone please help me with this
Mama L [17]

Answer:

27.

<em>Equation of 2 tangent lines at the given curve going through point P is:</em>

<em>y = -7x + 1</em>

<em>y = x + 1</em>

28.

<u>Part 1:</u>  400 feet

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

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

<u>Part 4:</u>  t = 10 seconds

29.

<u>Part 1:</u>  Average Rate of Change = -15

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

<u />

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

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

<em>y = -7x + 1</em>

<em>y = x + 1</em>

<em></em>

28.

<u>Part 1:</u>

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

<u>Part 2:</u>

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

<u>Part 3:</u>

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

<u>Part 4:</u>

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.

<u>Part 1:</u>

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

<u>Part 2:</u>

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
2 years ago
Can someone help me solve these two questions please
Vaselesa [24]

Answer:

hope this help

Step-by-step explanation:

sorry..I send pic instead..hope this help

4 0
2 years ago
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