No. It's rather simple, down to addition. Ask yourself, what does 6 + 8 equal? 6 + 8 = 14. Now is this an odd number? No, 14 is an even number.
I hope this helps! If not I'm sorry.
Well one similarity is that since you have to multiply, integer rules still apply and that you can also multiply variables. A difference is that with fractions, you have to make a number into an improper fraction rather than just the regular way which would most likely be the whole number with a decimal.
Answer:
d) 35ft
e) 6 seconds
f) 250°
Step-by-step explanation:
If it makes one rotation (360°) in 36 seconds, then it turns 10° every second.
36 secs = 360°
⇒ 1 sec = 360/36 = 10°
d) After 9 seconds, it has turned:
9 x 10 = 90°
Therefore, it will be 15 + 20 = 35ft from the wall
e) 60° = 6 seconds
f) 25 seconds = 25 x 10 = 250°
Answer:
x = 1 or x = -5/2
Step-by-step explanation:
Solve for x over the real numbers:
2 x^2 + 3 x - 2 = 3
Divide both sides by 2:
x^2 + (3 x)/2 - 1 = 3/2
Add 1 to both sides:
x^2 + (3 x)/2 = 5/2
Add 9/16 to both sides:
x^2 + (3 x)/2 + 9/16 = 49/16
Write the left hand side as a square:
(x + 3/4)^2 = 49/16
Take the square root of both sides:
x + 3/4 = 7/4 or x + 3/4 = -7/4
Subtract 3/4 from both sides:
x = 1 or x + 3/4 = -7/4
Subtract 3/4 from both sides:
Answer: x = 1 or x = -5/2
F - for simplification - we assume the orbits are circular (which they approximately are), we have a centrifugal force of <span><span>m<span><span>v2</span>R</span></span><span>m<span><span>v2</span>R</span></span></span><span>, where </span>m<span> is the mass of the planet, </span>v<span> its speed, and </span>R<span> the distance from the (center of the) sun. This force must be equal to the centripetal force, which comes from the sun's attraction, and it is </span><span><span>mG<span>M<span>R2</span></span></span><span>mG<span>M<span>R2</span></span></span></span><span>. Here, </span>M<span> is the mass of the sun, and </span>G<span> is a constant. We don't care about the actual numbers, so we write </span>C<span> be the product </span>MG<span> and observe that it is a constant independent of the orbit. These forces have to cancel out, so we have </span><span><span><span>v2</span>R=C<span>R<span>−2</span></span></span><span><span>v2</span>R=C<span>R<span>−2</span></span></span></span><span>, or </span><span><span><span>v2</span>=C<span>R<span>−3</span></span></span><span><span>v2</span>=C<span>R<span>−3</span></span></span></span><span>, so we have that the speed is proportional to </span><span><span>R<span><span>−3</span>2</span></span><span>R<span><span>−3</span>2</span></span></span><span>. Which means that if you decrease the distance to the sun, the speed goes up.</span>
