The Pythagorean theorem states that the square of the hypotenuse is equal to the sum of the squares of the legs:
The hypotenuse is the side with the greatest length.
Check all options:
1. c=5, a=3, b=4:
true.
2. c=14, a=4, b=11:
false.
3. c=17, a=9, b=14:
false.
4. c=16, a=8, b=14:
false.
5. c=17, a=8, b=15:
true.
Answer: correct options are A and E.
Answer:
∠A=65º
Step-by-step explanation:
- A angle of a line is 180º, so we can say that angle ABC=(180-6x)º
- The interior angkes of a triangle is 180º, so (x+40)º+(3x+10)º+(180-6x)º=180º
- Remove parenthesis, x+40+3x+10+180-6x=180
- Combine like terms, -2x+230=180
- Subtract 180, -2x=-50
- Divide by -2, so x=25
Now:
- ∠CAB=(25+40)º=65º
- ∠ABC=[180-6(25)]º=30º
- ∠BCA=[3(25)+10]=85º
Answer:
1 is B
2 is B
3 is stream line body and fluke both help in swimming and both are independent envolved that is their ancestors don't match
Step-by-step explanation:
Sadly, after giving all the necessary data, you forgot to ask the question.
Here are some general considerations that jump out when we play with
that data:
<em>For the first object:</em>
The object's weight is (mass) x (gravity) = 2 x 9.8 = 19.6 newtons
The force needed to lift it at a steady speed is 19.6 newtons.
The potential energy it gains every time it rises 1 meter is 19.6 joules.
If it's rising at 2 meters per second, then it's gaining 39.2 joules of
potential energy per second.
The machine that's lifting it is providing 39.2 watts of lifting power.
The object's kinetic energy is 1/2 (mass) (speed)² = 1/2(2)(4) = 4 joules.
<em>For the second object:</em>
The object's weight is (mass) x (gravity) = 4 x 9.8 = 39.2 newtons
The force needed to lift it at a steady speed is 39.2 newtons.
The potential energy it gains every time it rises 1 meter is 39.2 joules.
If it's rising at 3 meters per second, then it's gaining 117.6 joules of
potential energy per second.
The machine that's lifting it is providing 117.6 watts of lifting power.
The object's kinetic energy is 1/2 (mass) (speed)² = 1/2(4)(9) = 18 joules.
If you go back and find out what the question is, there's a good chance that
you might find the answer here, or something that can lead you to it.
