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3 years ago

A rectangle has a a perimeter of 72 ft. The length and width are scaled by a factor 3.5.

2 answers:

8 0

Formula for perimeter of rectangle:
P=2(L+w)
P=2L+2w
Here Perimeter, p= 72 ft
And scale factor of length and width=3.5 which means rectangle is enlarged.
So perimeter of rectangle= 3.5 x 72
P=252ft

Answer: 252 ft

Tasya [4]3 years ago

8 0

We know the perimeter of the original rectangle is 72ft.
We also know that the perimeter P of the rectangle is defined as:
P=2L+2W.
So P is directly proportional to L+W. This means that if L or W were to get bigger, P will get bigger, and if L or W were to get smaller P will get smaller.

The question also states that both L and W are scaled by the same factor of 3.5 .
This means the sum of the dimensions L+W is scaled by 3.5
It also means that the perimeter is upscaled by 3.5 as it is proportional to L+W.
P = 72 ft × 3.5 = 252 ft.

252 ft is the answer.

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3 years ago

How do you make a word problem for 36÷4

Flura [38]

Word Problem: There are 36 students in all. The teacher wants to make 4 groups. How much students go into each group?

Work: 36/4 = 9

Answer: 9 students go into each group

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3 years ago

The spread of a virus is modeled by V (t) = −t 3 + t 2 + 12t,

VashaNatasha [74]

Functions can be used to model real life scenarios

The reasonable domain is $\mathbf{[0,\infty)}$ .
The average rate of change from t = 0 to 2 is 20 persons per week
The instantaneous rate of change is $\mathbf{V'(t) = -3t^2 + 2t + 12}$ .
The slope of the tangent line at point (2,V(20) is 10
The rate of infection at the maximum point is 8.79 people per week

The function is given as:

$\mathbf{V(t) = -t^3 + t^2 + 12t}$

(a) Sketch V(t)

See attachment for the graph of $\mathbf{V(t) = -t^3 + t^2 + 12t}$

(b) The reasonable domain

t represents the number of weeks.

This means that: t cannot be negative.

So, the reasonable domain is: $\mathbf{[0,\infty)}$

(c) Average rate of change from t = 0 to 2

This is calculated as:

$\mathbf{m = \frac{V(a) - V(b)}{a - b}}$

So, we have:

$\mathbf{m = \frac{V(2) - V(0)}{2 - 0}}$

$\mathbf{m = \frac{V(2) - V(0)}{2}}$

Calculate V(2) and V(0)

$\mathbf{V(2) = (-2)^3 + (2)^2 + 12 \times 2 = 20}$

$\mathbf{V(0) = (0)^3 + (0)^2 + 12 \times 0 = 0}$

So, we have:

$\mathbf{m = \frac{20 - 0}{2}}$

$\mathbf{m = \frac{20}{2}}$

$\mathbf{m = 10}$

Hence, the average rate of change from t = 0 to 2 is 20

(d) The instantaneous rate of change using limits

$\mathbf{V(t) = -t^3 + t^2 + 12t}$

The instantaneous rate of change is calculated as:

$\mathbf{V'(t) = \lim_{h \to \infty} \frac{V(t + h) - V(t)}{h}}$

So, we have:

$\mathbf{V(t + h) = (-(t + h))^3 + (t + h)^2 + 12(t + h)}$

$\mathbf{V(t + h) = (-t - h)^3 + (t + h)^2 + 12(t + h)}$

Expand

$\mathbf{V(t + h) = (-t)^3 +3(-t)^2(-h) +3(-t)(-h)^2 + (-h)^3 + t^2 + 2th+ h^2 + 12t + 12h}$ $\mathbf{V(t + h) = -t^3 -3t^2h -3th^2 - h^3 + t^2 + 2th+ h^2 + 12t + 12h}$

Subtract V(t) from both sides

$\mathbf{V(t + h) - V(t)= -t^3 -3t^2h -3th^2 - h^3 + t^2 + 2th+ h^2 + 12t + 12h - V(t)}$

Substitute $\mathbf{V(t) = -t^3 + t^2 + 12t}$

$\mathbf{V(t + h) - V(t)= -t^3 -3t^2h -3th^2 - h^3 + t^2 + 2th+ h^2 + 12t + 12h +t^3 - t^2 - 12t}$

Cancel out common terms

$\mathbf{V(t + h) - V(t)= -3t^2h -3th^2 - h^3 + 2th+ h^2 + 12h}$

$\mathbf{V'(t) = \lim_{h \to \infty} \frac{V(t + h) - V(t)}{h}}$ becomes

$\mathbf{V'(t) = \lim_{h \to \infty} \frac{ -3t^2h -3th^2 - h^3 + 2th+ h^2 + 12h}{h}}$

$\mathbf{V'(t) = \lim_{h \to \infty} -3t^2 -3th - h^2 + 2t+ h + 12}$

Limit h to 0

$\mathbf{V'(t) = -3t^2 -3t\times 0 - 0^2 + 2t+ 0 + 12}$

$\mathbf{V'(t) = -3t^2 + 2t + 12}$

(e) V(2) and V'(2)

Substitute 2 for t in V(t) and V'(t)

So, we have:

$\mathbf{V(2) = (-2)^3 + (2)^2 + 12 \times 2 = 20}$

$\mathbf{V'(2) = -3 \times 2^2 + 2 \times 2 + 12 = 4}$

Interpretation

V(2) means that, 20 people were infected after 2 weeks of the virus spread

V'(2) means that, the rate of infection of the virus after 2 weeks is 4 people per week

(f) Sketch the tangent line at (2,V(2))

See attachment for the tangent line

The slope of this line is:

$\mathbf{m = \frac{V(2)}{2}}$

$\mathbf{m = \frac{20}{2}}$

$\mathbf{m = 10}$

The slope of the tangent line is 10

(g) Estimate V(2.1)

The value of 2.1 is

$\mathbf{V(2.1) = (-2.1)^3 + (2.1)^2 + 12 \times 2.1}$

$\mathbf{V(2.1) = 20.35}$

(h) The maximum number of people infected at the same time

Using the graph, the maximum point on the graph is:

$\mathbf{(t,V(t) = (2.361,20.745)}$

This means that:

The maximum number of people infected at the same time is approximately 21.

The rate of infection at this point is:

$\mathbf{m = \frac{V(t)}{t}}$

$\mathbf{m = \frac{20.745}{2.361}}$

$\mathbf{m = 8.79}$

The rate of infection is 8.79 people per week

Read more about graphs and functions at:

brainly.com/question/18806107

6 0

3 years ago

Mr. Smith wants to order some pizza and cheese pizza cost $11 and pepperoni pizza cost $12.50 if you bought 13 pieces total and

Triss [41]

Answer: Mr Smith bought 7 pepperoni pizza

Step-by-step explanation:

Step 1

cost of cheese pizza =$11

pepperoni pizza cost =$12.50

let number cheese pizza be represented as c

and number of pepperoni pizza be represented as p

such that the total number of pizza bought which equals 13 can be represented as

c+ p= 13

and total cost of both pizza can be expressed as

11c+ 12.50p =153.50

Step 2 Solving

c+ p= 13

11c+ 12.50p =153.50

By elimination method, Multiply equation 1 by 11 and subtract the new equation from equation 2

11c+ 12.50p =153.50

11c+ 11p=143.

1.5p =10.5

p= 10.5/ 1.5 =7

c+ p= 13

c= 13-p =13-7=6

Therefore Mr Smith bought 6 cheese pizza and 7 pepperoni pizza

6 0

3 years ago

Find the equation of a straight line passing throught the point listed and having the given gradient. Express your answer in the

Igoryamba

First let's try to find the equation in this form : y = mx + c

The gradient is given 3 . In a line's equation, x's coefficient represents the line's gradient.

So equation of a line with the gradient of 3, would look like this ;

 $y= 3x + c$

Now a point that the line passes through is given, (1, 2)

This point's x-coordinate is 1 and y-coordinate is 2.

So we'll plug its x-coordinate value in the equation and also y-coordinate value. So we can solve it.

As you know, $x=1$ and $y=2$

$y = 3x + c$

$2\quad =\quad 3\cdot 1+c\\ \\ 2\quad =\quad 3+c\\ \\ 2-3\quad =\quad c\\ \\ -1\quad =\quad c$

We found c = -1

Also in a line's equation, c is constant and it represents the line's y-intercept

So let's build the line's equation.

$m=3$ and $c=-1$

$y= mx + c$

$y= 3x -1$

We found the line's equation in this form, $y= mx + c$

Now let's turn it into this form, $ax + by + c = 0$

$y\quad =\quad 3x-1\\ \\ y-3x\quad =\quad -1\\ \\ y-3x+1\quad =\quad 0\\ \\ -3x+y+1\quad =\quad 0$

Final answers,

$\boxed { y\quad =\quad 3x-1 }$

and

$\boxed { -3x+y+1\quad =\quad 0 }$

I hope this was clear enough :)

5 0

4 years ago

Read 2 more answers