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Arte-miy333 [17]
3 years ago
7

There was 1/5 of a gallon of lemonade in the refrigerator. Kelly poured equal amounts into 4 cups . How much lemonade was in eac

h cup ?
Mathematics
1 answer:
3241004551 [841]3 years ago
3 0

Answer:

1/20 gallon per cup

Step-by-step explanation:

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How many terms in 9a+11p
Naddik [55]

Answer:

2 terms

Step-by-step explanation:

Terms are nothing but the mixtures of co-effecients and their variables..

Here, 9a is one term while 11p is the other

7 0
3 years ago
Read 2 more answers
Analyze the diagram below and complete the instructions to follow.<br><br> Find a, b and c
Feliz [49]

Answer/Step-by-step explanation:

✔️Using trigonometric ratio, find b:

Reference angle = 45

opposite = 8

Hypotenuse = b

Thus,

Sin(45) = 8/b

b = 8/sin(45)

b = 8/(1/√2) (sin 45 = 1/√2)

b = 8 × √2/1

b = 8√2

✔️Find a using trigonometric ratio:

Reference angle = 60°

Hypotenuse = b = 8√2

Adjacent = a

Therefore,

Cos(60) = a/8√2

8√2*cos(60) = a

8√2*½ = a

4√2 = a

a = 4√2

✔️Find c using Pythagorean Theorem:

c² = b² - a²

c² = (8√2)² - (4√2)²

c² = 128 - 32

c² = 96

c = √96

c = √(16*6)

c = 4√6

5 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
Cornerstone Bakery sold 78 pies on Monday 96 pies on Tuesday 40 pies on Wednesday 104 pies on Thursday and 77 pies on Friday on
just olya [345]
They sold an average of 79 pies a day
6 0
3 years ago
|x+2|+4=11<br> A) x=5,-9
Lisa [10]

Answer:

the mod always gives positive answers

so the answer will be

Step-by-step explanation:

x+2+4=11

x+6=11

x=11-6

x=5

answer is x is 5

8 0
3 years ago
Read 2 more answers
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