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WARRIOR [948]
3 years ago
5

5 pounds for $19.00 Cost per pound how much is it for one pound​

Mathematics
2 answers:
MAVERICK [17]3 years ago
6 0

Answer:

The cost for one pound would be 3.8

Step-by-step explanation:

if you do 19divided by 5 you get 3.8

jeka57 [31]3 years ago
3 0

HUNTER X HUNTER

wrong ANIME gtg

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Three U.F.O.'s were constructed in consecutive years. Their ages have a sum of
Dahasolnce [82]

Answer: 41, 42, and 43 years old.

Step-by-step explanation:

41 plus 42 plus 43 = 126.

7 0
3 years ago
Find the 11th term of the arithmetic sequence an=<br> 24 – 7(n - 2)<br> 2115
sweet [91]

Answer:

Let's simplify

24−7(n−2)(2115)

Step-by-step explanation:

step-by-step.

=−14805n+29634

I hope my answer helped

7 0
2 years ago
Greta has watched 30 minutes of a movie. This is 20%<br>of the entire movie. How long is the movie?​
german
The answer is 180 minutes. because you multiply 30 times 0.20 and you get 6. Then you multiply 6 times 30 and get 180 minutes. And 180 minutes in hours is 3 hours.
8 0
3 years ago
Dilution Problem
LUCKY_DIMON [66]

Answer:

  76.7 liters

Step-by-step explanation:

You have ...

  C1×V1 = C2×V2

so ...

  V2 = V1×(C1/C2) = (115 L)×(14/21) = 76 2/3 L

  V2 ≈ 76.7 L

4 0
3 years ago
An acute triangle has sides measuring 10 cm and 16 cm. The length of the third side is unknown.
saveliy_v [14]

Answer: Choice B

12.5 < x < 18.9

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

Explanation:

We have a triangle with these side lengths:

  • a = 10
  • b = 16
  • c = x = unknown

Let's assume that b = 16 is the largest side of this triangle.

By the converse of the pythagorean theorem, we need b^2 < a^2+c^2 to be true in order for an acute triangle to happen.

So,

b^2 < a^2 + c^2\\\\c^2 > b^2 - a^2\\\\c > \sqrt{b^2-a^2}\\\\x > \sqrt{16^2-10^2}\\\\x > \sqrt{156}\\\\x > 12.4899959967968 \ \text{(approximate)}\\\\x > 12.5

Now let's consider the possibility that the missing side x is actually the longest side.

Using the same theorem as before, we would say,

c^2 < a^2 + b^2\\\\c < \sqrt{a^2 + b^2}\\\\x < \sqrt{10^2 + 16^2}\\\\x < \sqrt{356}\\\\x < 18.8679622641132 \ \text{(approximate)}\\\\x < 18.9\\\\

We found that x > 12.5 and x < 18.9

This is the same as saying 12.5 < x and x < 18.9

Put together, they form the approximate answer of 12.5 < x < 18.9

6 0
2 years ago
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