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lana66690 [7]
3 years ago
15

I WILL AWARD BRAINLIEST!!!! PLEASE HELP!!!!!!

Mathematics
2 answers:
MrRa [10]3 years ago
5 0

Answer:

1  4(3-r)  or  12-4r

2.  7( 8-m) or 56-7m

3.   -2

Step-by-step explanation:

1) |4r−12| , if r<3

|4r-12|

Factor out a 4

4|r-3|

|r-3| will be less than 0 since r is less than 3

So we can rewrite it as 3-r to remove the absolute value signs and make it positive

4 (3-r)

12 -4r

2.   |7m–56| , if m<8

Factor out a 7

7| m-8|  

|m-8| will be less than 0 since m is less than 8

So we can rewrite it as  8-m to remove the absolute value signs and make it positive

7( 8-m)

56-7m

3) |z−7|−|z−9| , if z<7

|z-7| will be less than 0 since z is less than 7  and |z-9| will be less than 0 since z is less than 7

So we can rewrite it as  7-z  and 9-z to remove the absolute value signs and make it positive

7-z - (9-z)

Distribute the minus sign

7-z-9+z

-2

yarga [219]3 years ago
3 0

Answer:

1  4(3-r)  or  12-4r

2.  7( 8-m) or 56-7m

3.   -2

Step-by-step explanation:

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What is the volume of soup that will fit in a cylindrical-shaped can with a height that is 2 inches longer than the radius?
trasher [3.6K]

Answer:

V = πr³ + 2πr²

Step-by-step explanation:

Volume of a cylinder:

V = πr²h

with radius r and height h = 2 + r:

V = πr²(2 + r) = πr³ + 2πr²

4 0
3 years ago
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City A is 10 miles from city B and 5 miles from city C. Cities A, B, and C form a right triangle at A. A road connects cities B
laila [671]

Answer:

11.18 miles

Step-by-step explanation:

The sides of a right triangle a, b, c are related by the equation

a^2 + b^2 = c^2

where c is the longest side of the triangle

Hence given that A connects B and C directly and forms a right angle at A

5^2 + 10^2 = c^2

Where c is the length of the road

25 + 100 = c^2

125 = c^2

c = 125^1/2

= 11.18 miles

3 0
2 years ago
I need help on this question
gayaneshka [121]

Answer:

the answer is letter D you reflex over the x axis which is he middle line, so the shape will be mirrored, then you would translate it to the right (move to the right) to be in the same place as Z

3 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 />

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3 years ago
HALP MEH
Mumz [18]
<span>A peer-viewed article. </span>
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