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Vlada [557]
2 years ago
6

There are 20,000 owls in the wild. Every decade, the number of owls halved. is this linear or Exponential.

Mathematics
1 answer:
erica [24]2 years ago
8 0

Answer:

This is exponential

Step-by-step explanation:

Hope this helps! Have a nice day!

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Angle XYZ is rotated 270 degrees counterclockwise about the origin to form angle X’Y’Z’. Which statement shows the measure of an
Kruka [31]
Hi i a doing the same thing and i have the answer as a hope this helps
6 0
3 years ago
Coach Pettaway is buying basketball equipment for his team. He gets a
soldier1979 [14.2K]
First, let’s all acknowledge that whoever comes up with problems like this WANTS kids to hate math...smh

I’m sure there is a prettier way to solve this, but here’s what I did:

8(2.25) + 3(22.50) =
18 + 67.50 = 85.50 per “set” of balls/jerseys
400/85.50 = 4.678 = number of “sets” he can buy. Round down to 4 so we have room for tax.

85.5 x 4 “sets”= $342
Tax on 342 is 0.06 x 342 = 20.52

$342 + 20.52 = $362.52 spent

Basketballs = 4 sets x 8 balls per set= 32
Jerseys = 4 sets x 3 jerseys per set= 12

32 basketballs, 12 jerseys, $362.52 spent

3 0
3 years ago
ΔCDE is reflected over the x-axis. What are the vertices of ΔC'D'E' ?
sattari [20]

Answer:

B.

Step-by-step explanation:

When reflecting over the x-axis:

(x, y) (x, -y)

The y changes signs (+, -)

7 0
2 years ago
The steps to convert fractions into decimals
Allisa [31]

Answer:

Just simplify!!

EXample: 11 over 8  

11

8

=

1 and 3 over 8  

1

3

8

=1.375

7 0
3 years ago
Suppose that five ones and four zeros are arranged around a circle. Between any two equal bits you insert a 0 and between any tw
PolarNik [594]

Answer:

Using <u>backward reasoning</u> we want to show that <em>"We can never get nine 0's"</em>.

Step-by-step explanation:

Basically in order to create nine 0's, the previous step had to have all 0's or all 1's. There is no other way possible, because between any two equal bits you insert a 0.

If we consider two cases for the second-to-last step:

<u>There were 9 </u><u>0's</u><u>:</u>

We obtain nine 0's if all bits in the previous step were the same, thus all bit were 0's or all bits were 1's. If the previous step contained all 0's, then we have the same case as the current iteration step. Since initially the circle did not contain only 0's, the circle had to contain something else than only 0's at some point and thus there exists a point where the circle contained only 1's.

<u>There were 9 </u><u>1's</u><u>:</u>

A circle contains only 1's, if every pair of the consecutive nine digits is different. However this is impossible, because there are five 1's and four 0's (we have an odd number of bits!), thus if the 1's and 0's alternate, then we obtain that 1's that will be next to each other (which would result in a 1 in the next step). Thus, we obtained a contradiction and thus assumption that the circle contains nine 0's after iteratins the procedure is false. This then means that you can never get nine 0's.

To summarize, in order to create nine 0's, the previous step had to have all 0's or al 1's. As we didn't start the arrange with all 0's, the only way is having all 1's, but having all 1's will not be possible in our case since we have an odd number of bits.

<u />

5 0
3 years ago
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