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Mrrafil [7]
2 years ago
13

I will give you a Brainliest

Mathematics
1 answer:
vladimir2022 [97]2 years ago
8 0

if the circumference of the shield is 100.48inches, nits radius is 16 inches and the area of the front of the shield is about 803.84 square inches.

proof

circumference = 2x r x Pi implies r =circumference / 2Pi=100.48/2x3.14=16

area = Pi x r²=3.14 x 16²=804.84

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Which of these ordered pairs is the best estimate of the solution of the system y = 3x + 2 and y = -1/4 x + 1?
Lelu [443]
If the solution exists, y=y....

3x+2=-x/4+1  make everything have a common denominator of 4...

(12x+8)/4=(-x+4)/4 now multiply both sides by 4

12x+8=4-x  add x to both sides

13x+8=4  subtract 8 from both sides

13x=-4  divide both sides by 13

x=-4/13

x≈-0.31  and the only pair in your choices having an x value close to that is:

(-0.3, 1.1)
6 0
3 years ago
Tickets to a museum cost $17 each. for a field trip, the museum offers a $4 discount on each ticket. how much will tickets for 3
My name is Ann [436]
I hope this helps you

4 0
3 years ago
WILL MARK U AS BRAINLIEST!!! Evaluate the following expressions given the functions below:
Scorpion4ik [409]

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7 0
3 years ago
Find the length of UC on the attached diagram. Thank you!
ValentinkaMS [17]
<h2>Answer </h2>

The length of UC is 18

<h2>Explanation </h2>

First we are going to find the length of JN; then we are subtracting from it the length of JU plus the length of CN.

We can infer from our picture that JN is 82 + 105, so JN = 187

We can also infer that JU = JH + HU

                                     JU = 22 + 96

                                     JU = 118

We can also infer that CN = 51

Now we can fin the length of UC:

UC=JN-(JU+CN)

UC=187-(118+51)

UC=18

We can conclude that the length of UC is 18.

3 0
3 years ago
Read 2 more answers
Find the number of ways of arranging the numbers
Doss [256]

First of all, note that all integers are either 0,1, or 2 modulo 3 (if you're not familiar with this terminology, it means that every integer is either a multiple of 3, or it is 1 or 2 away from a multiple of 3).

So, we can think of our numbers as

\begin{array}{c|c}x&x\mod 3\\0&0\\1&1\\2&2\\3&0\\4&1\\5&2\\6&0\\7&1\\8&2\\9&0\end{array}

In order to make sure that the sum of any three adjacent numbers is divisible by 3, we have to make sure that any group of 3 three adjacent numbers contains a 0, a 1 and a 2. This is possible only if we arrange our 9 numbers in 3 groups of 3 numbers containing 0,1 and 2 exactly once, repeating always the same pattern.

For example, we could arrange our numbers following the pattern

0,1,2,0,1,2,0,1,2

or

2,0,1,2,0,1,2,0,1

We have 3!=6 possible patterns. Suppose for example that we choose the pattern

0,1,2,0,1,2,0,1,2

One possible way of following this pattern would be the arrangement

3,1,2,6,4,5,9,7,8

In fact, we substituted every '0' with a multiple of 3 (3, 6 or 9), every '1' with a number 1 away from a multiple of 3 (1, 4 or 7) and every '2' with a number 2 away from a multiple of 3 (2, 5 or 8).

This means that, once we fix a patter, we have 3 choices for the first 3 slots, 2 choices for the next 3 slots, and the final slot will be fixed. So, we have

3\cdot 3\cdot 3\cdot 2 \cdot 2 \cdot 2 = 216

possible ways of following a fixed pattern. Since the number of patterns was 6, we have

216\cdot 6 = 1296

possible arrangements.

7 0
3 years ago
Read 2 more answers
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