You have to combine like terms, so the variable (x, y, s, d, c....) and the exponents must be the same in order to combine them.
For example:
x² + x³ Since they don't have the same exponent, you can't combine them
y² + 3y² = 4y²
23x + x = 24x
4. 2s² + 1 + s² - 2s + 1 You can rearrange it if it makes it easier
2s² + s² - 2s + 1 + 1 = 3s² - 2s + 2
5. 5t² - 2t - 1 - (3t² - 5t + 7) Distribute/multiply the - to (3t² - 5t + 7)
5t² - 2t - 1 - 3t² + 5t - 7 = 2t² + 3t - 8
Do the same for #9 and #10, and you should get:
9. 2k² + 5k - 9
10. 6y³ - 7y² - 6y - 12
Since C is 2mi east of A and 2mi south of B, the shortest distance from A to B would be a straight line which is equal to the hypotenuse of a right triangle with a base of 2mi and a height of 2mi..
d^2=x^2+y^2
d^2=2^2+2^2
d^2=4+4
d^2=2*4
d=√(2*4)
d=2√2 mi (exact)
d≈2.83 mi (to nearest hundredth of a mile)
Answer: b is 6
Step-by-step explanation:
Set the proportions up and solve for b
b. 3
---- = ------
10. 5
Ok so we need to set the bottom to 0 to find the vertical asympyotes. This becomes x^2 - 4 = 0. Since we're talking about asymptotes, i'll assume you can solve basic equations. Solving for x and you get x = ±2. This means the vertical asymptotes are at ±2. To solve for horizontal asymptotes you take the limit as x goes to ±∞. Either way you end up with ±∞/∞. Now this isn't 1 because they grow at different rates. You differentiate both the top and bottom(L'hopital) and you get 6x/2x which becomes 3. This means the horizontal asymptote is at y = 3.
