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Marrrta [24]
3 years ago
7

Looking for the Domain and range

Mathematics
1 answer:
Nuetrik [128]3 years ago
3 0

Answer:

Its ]-infinity,+infinity[ for both domain and range

Step-by-step explanation:

You might be interested in
2u^3 v^2<br> _______<br> (2u)^3 *u^2
Rus_ich [418]

Answer:

v^2 over 4u^2

Step-by-step explanation:


4 0
3 years ago
Please help me to get correct answer
valentinak56 [21]

Answer:

1. Because a right angle is 90 degrees, the answer to this is 90 - 65 = 25.

2. This question is similar to the last one, but this time the angle is 180 degrees total, so the answer is 180 - 55 = 125.

3. I'm fairly certain the lines here indicate that all three angles are the same. If all three angles make 180 degrees, the answer would be 180 / 3 = 60.

4. This is the same as in question 2. 180 - 135 = 45.

5. Again, because all three angles need to make up 180 degrees, we can just simply do 180 - (65 + 45) = 70.

6. With a quadrilateral, the angles add up to 360 degrees. We can do the same thing as in question 5 here. 360 - (70 + 95 + 55) = 140.

7. I don't remember exactly how to solve this but 55 degrees is correct because 55 + 55 + 70 = 180 which is the only possible value.

I haven't done this kind of stuff in a looooong time so I hope this suffices! :)

7 0
3 years ago
What is the a3+b3 equal to? verify by an activity
uranmaximum [27]
a³ + b³ = ( a + b ) ( a² - a b + b³ )
For example:
a = 1, b = 2
a³ + b³ = 1³ + 2³ = 1 + 8 = 9
( a + b ) ( a² - a b + b² ) = ( 1 + 2 ) ( 1² - 1 * 2 + 2² ) =
= 3 * 3 = 3
The formula is true. 
3 0
3 years ago
If I paid $9.00 for 4 lb of candies, how much candies can I buy for $12.00?
Lina20 [59]

Answer:

We are given that 4 lb of candies are bought for $9.00

Therefore, the price of 1 lb of candies is:

\frac{9}{4}=2.25

Now the amount of candies that can bought for $12.00 is:

\frac{12}{2.25} =5.3333 lb

Therefore, 5.3333 lb of candies can be bought for $12.00


3 0
3 years ago
**PLEASE HELP ASAP**
adell [148]
In June he read 0 to 45 pages per day
In April he reads 10 to 90 pages per day
7 0
2 years ago
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