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Blizzard [7]
3 years ago
9

A rectangular piece of paper with length 35 cm and a width 18 cm has two semicircles cut out of it as shown below

Mathematics
1 answer:
Eddi Din [679]3 years ago
3 0

Answer:

375.66 cm²

Step-by-step explanation:

The two semicircles add to a full circle.

For the circle, d = 18 cm.

r = d/2 = 9 cm.

A = LW - πr²

A = (35 cm)(18 cm) - 3.14(9 cm)²

A = 375.66 cm²

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Plzz I asked this three times! The Functions r and s are defined as follows.
Lana71 [14]

Answer:

s(r(2)) = 6

Step-by-step explanation:

You would first solve for r(2). You then input r(2) which equals -2 into s(r(2)).

r(x)= -2x + 2

r(2)= -2(2) + 2

r(2)= -4 +2

r(2)= -2

s(x)=x² + 2

s(r(2)) = (-2)² +2

s(r(2)) = 6

6 0
3 years ago
ITS been increased by 39%​
Alenkasestr [34]

Answer:

Step-by-step explanation:

Let 'x' be the original price.

x + 39% of x = 197.38

x + 0.39 x   = 197.38

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Converting measures 48ft= yd​
Vera_Pavlovna [14]

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48 Feet (ft) 16.000 Yards (yd)

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Step-by-step explanation:

5 0
3 years ago
Plzz anyone solve all answers plzzzzzzzz​
algol13

You posted a lot of problems here. In the future please only post one problem at a time. Thank you.

I'll do the first two problems to get you started. Hopefully it will help you finish off the rest of the questions.

==========================================

Problem 1

{18, a, b, -3} is an arithmetic sequence or arithmetic progression (AP).

This means we have some number d added on to each term to get the next term.

first term = 18

second term = first term + d = 18+d = a

third term = second term + d = (18+d)+d = 18+2d = b

fourth term = third term + d = (18+2d)+d = 18+3d = -3

----

Let's solve that last equation for d

18+3d = -3

18+3d-18 = -3-18

3d = -21

3d/3 = -21/3

d = -7

----

The value d = -7 tells us to add -7 to each term to get the next term. In other words, we subtract 7 from each term to get the next term

first term = 18

second term = first term + d = 18+d = 18+(-7) = 18-7 = 11

third term = second term + d = 11+d = 11+(-7) = 11-7 = 4

fourth term = third term + d = 4+d = 4+(-7) = 4-7 = -3

----

We see that a = 11 and b = 4 are the second and third terms respectively.

Therefore, a+b = 11+4 = 15

-------------

<h3>Answer: 15</h3>

==========================================

Problem 2

A multiple of 4 is in the form 4*n for some integer n, ie n is a whole number.

We want to know which values of 4*n are between 10 and 250.

----

Divide both 10 and 250 by 4 to get the following

10/4 = 2.5

250/4 = 62.5

If n = 2, then 4*n = 4*2 = 8 is not between 10 and 250; however n = 3 will make 4*n = 4*3 = 12 to be between 10 and 250. We see that n = 3 is the smallest possible allowed value.

If n = 62, then 4*n = 4*62 = 248 is between 10 and 250; while n = 63 will make 4*n too big because 4*63 = 252. The largest n can get is n = 62

----

The question posed in question 2 is equivalent to asking the following: "How many values are in the set {3, 4, 5, ..., 60, 61, 62}?"

You could count all of the values in the set, but that exercise is very tedious busywork. There's a much faster way. First lets consider the set below

{a, a+1, a+2, ..., b-2, b-1, b}

where a,b are integers. Basically this set starts at 'a', counts up until we get to 'b'. The handy formula

c = b-a+1

will provide the exact count of values in the set {a, a+1, a+2, ..., b-2, b-1, b}

----

In this case, a = 3 and b = 62, making

c = b-a+1

c = 62-3+1

c = 60

There are 60 values in the set {3, 4, 5, ..., 60, 61, 62}

There are 60 multiples of four that are between 10 and 250.

-------------

<h3>Answer: 60</h3>
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Answer:

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Step-by-step explanation:

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