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DaniilM [7]
3 years ago
12

Question 1 of 5

Mathematics
1 answer:
dedylja [7]3 years ago
5 0

Formula for the circumference of a circle:

C = πd    or     C = 2πr      

[C = circumference    π(pi)       d = diameter        r = radius(half the diameter)]

You know:

d = 5 feet        

and you're using 3.14 for π, so substitute/plug it into the equation

C = πd

C = (3.14)(5)

C = 15.7 feet       Your answer is B

You might be interested in
Graph for f(x)=6^6 and f(x)=14^x
zlopas [31]

Graph Transformations

There are many times when you’ll know very well what the graph of a

particular function looks like, and you’ll want to know what the graph of a

very similar function looks like. In this chapter, we’ll discuss some ways to

draw graphs in these circumstances.

Transformations “after” the original function

Suppose you know what the graph of a function f(x) looks like. Suppose

d 2 R is some number that is greater than 0, and you are asked to graph the

function f(x) + d. The graph of the new function is easy to describe: just

take every point in the graph of f(x), and move it up a distance of d. That

is, if (a, b) is a point in the graph of f(x), then (a, b + d) is a point in the

graph of f(x) + d.

As an explanation for what’s written above: If (a, b) is a point in the graph

of f(x), then that means f(a) = b. Hence, f(a) + d = b + d, which is to say

that (a, b + d) is a point in the graph of f(x) + d.

The chart on the next page describes how to use the graph of f(x) to create

the graph of some similar functions. Throughout the chart, d > 0, c > 1, and

(a, b) is a point in the graph of f(x).

Notice that all of the “new functions” in the chart di↵er from f(x) by some

algebraic manipulation that happens after f plays its part as a function. For

example, first you put x into the function, then f(x) is what comes out. The

function has done its job. Only after f has done its job do you add d to get

the new function f(x) + d. 67Because all of the algebraic transformations occur after the function does

its job, all of the changes to points in the second column of the chart occur

in the second coordinate. Thus, all the changes in the graphs occur in the

vertical measurements of the graph.

New How points in graph of f(x) visual e↵ect

function become points of new graph

f(x) + d (a, b) 7! (a, b + d) shift up by d

f(x) Transformations before and after the original function

As long as there is only one type of operation involved “inside the function”

– either multiplication or addition – and only one type of operation involved

“outside of the function” – either multiplication or addition – you can apply

the rules from the two charts on page 68 and 70 to transform the graph of a

function.

Examples.

• Let’s look at the function • The graph of 2g(3x) is obtained from the graph of g(x) by shrinking

the horizontal coordinate by 1

3, and stretching the vertical coordinate by 2.

(You’d get the same answer here if you reversed the order of the transfor-

mations and stretched vertically by 2 before shrinking horizontally by 1

3. The

order isn’t important.)

74

7:—

(x) 4,

7c’

‘I

II

‘I’

-I

5 0
2 years ago
GIVING AWAY 50 POINTS TO WHOEVER SHOWS LEGIT WORK!
miss Akunina [59]
S= one length of square 1
S+1 = one length square 2
S+2= one length square 3

Our equation would be
S^2+(S+1)^2+(S+2)^2=365

Idk if it would be called distribute but.. distribute
S^2+S^2+2S+1+S^2+4S+4=365

Add like terms
3S^2+6S-360=0

Divide everything. By 3
S^2+2S-120=0

Factor
(S-10)(S+12)=0

The two solutions are:

S-10=0
S=10

S+12=0
S=-12

Since a length can’t be a negative the only possible solution would be 10

Since a perimeter is all lengths added together we can multiply the length by four to get the perimeter

Square 1
10*4=40
Perimeter is 40cm

Square 2
S+1 =11
11*4=44
Perimeter is 44cm

Square 3
S+2=12
12*4=48
Perimeter is 48cm

Add all the perimeters together to get the total perimeter:

Total perimeter:
40+44+48=132

The total perimeter is 132cm

I hope this helps. Sorry if I messed up anything on here it was kinda hard to keep track of everything. Feel free to ask if you need anything cleared up :)
6 0
3 years ago
Choose the option that best answers the question. Paracelsus University has two kinds of professors: academic professors and pro
NARA [144]

Answer:

percentage of professional professor are 75

Step-by-step explanation:

Given data

academic professors A = 60%

professors  tenured P = 70%

professors at Paracelsus University = 90%

to find out

what percent of the professional professors

solution

we know 90% of the professors are academic professors or tenured or both so we can say percent of academic professors = 60 + 70 - 90 = 40

because here

total = A + P - both

90 = 60 + 70 - both

both = 40

so professors  tenured will be here 70 - 40 = 30

so

percentage of professional professor are = 30 / 40 × 100

percentage of professional professor are 75

5 0
2 years ago
Rewrite the equation by completing the square.
Dennis_Churaev [7]
Once you complete the square, the simplified version of this equation is

(2x - 1)^2= 0

If you want to solve for x, subtract 1 and divide by 2.

x = 0.5
7 0
3 years ago
Help me out pleaseee !!!
ioda
What do you neeeeeed help onnn i got youuuuu just tell meeeeeee
4 0
3 years ago
Read 2 more answers
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