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Rom4ik [11]
4 years ago
6

1. The people of Whoville were shopping like crazy to get their lists completed! one customer bought a wreath for $12 and seven

ornaments. The customer spent a total of $54. How much did each ornament cost?
2. If the average size of a heart is 12cm and the Grinch's heart is two sizes too small, How many centimeters is the Grinch's heart?
3. The Grinch goes up the mountain in the Dump-It-To-Grump-It which is a total of 15 miles from Whoville. The last bump of the Dump is 5 miles away. How many total miles were the previous bumps up the mountain?
4. The Grinch X-Rays his heart and it is down a size and a half. If his heart measured 20cm, to begin with, what is the current measure?
5. How old was the Grinch when his classmates picked on him? He is 150 now and if I reduce his current age by three times his age as a kid, it would be 126.
6. The Grinch asked for a three-quarter-inch wrench. How is this written as a percent?
7. The Grinch has termites in his smile. If the Grinch has a total of 55 teeth and the termites have infested 20% of his teeth, how many teeth are infested with termites?
8. The Grinch made a belt buckle that is rectangular shaped. The buckle has a length of 7 inches and a width of 3 inches. What is the total area of the belt buckle?
9. When the Grinch arrives at his first house, he takes all the decorations. By mistake, he has left 8 ornaments on the floor. If this is 2% of the total number of ornaments, how many ornaments were there to start with?
10. The Grinch twists one light bulb which connects to all the lights and they all turn back on. If this one light represents one percent of the total lights on the strand, how many total lights were on the strand?
11. There are a lot of Whos sitting at the long table having the Christmas feast. If the Grinch and his dog represent 5% of the total number of Whos sitting at the table, how many total Whos (including the Grinch and his dog) are enjoying the Christmas Feast?
Mathematics
1 answer:
OlgaM077 [116]4 years ago
4 0
1 = 6
5=8
6=75%?
7=21?
8=21
9=400
10=100
11=40

These are all the one I got the ones with ? At the end I’m not sure about
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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
3 years ago
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