Answer:
AC is about 41. 095 miles long.
Step-by-step explanation:
You can use the Law of Cosines to solve this: c^2 = a^2 + b^2 – 2ab cos(C)
Substitute in your values, and you're good to go!
c^2=16^2+28^2-2(16)(28)(cos(117))
c^2 = 256 + 784 - 896(cos(117))
c = sqrt(1040 - 896(cos(117)))
c is about 41. 095 miles long.
Answer:
A. No, this is not a valid inference because he asked only 20 families
Step-by-step explanation:
You cannot assume that those are the only families that would want to go; you have to ask all the families.
Green 29. white 37 black 32
My answer -
<span>1. Use symbols (not words) to express quotient
2. Use exponent symbol (^) to denote exponents
3. Just write out question number, question, and choices. No need for
extra information (such as points). Also, don't leave blank lines
between choices. This extraneous that we don't need just makes your
whole question very very long, and means a lot of scrolling on our part.
4. You should only post 2 or 3 questions at a time.
1) (6x^3 − 18x^2 − 12x) / (−6x) = −x^2 + 3x + 2 ----> so much simpler to read !
2) (d^7 g^13) / (d^2 g^7) = d^(7−2) g^(13−7) = d^5 g^6 ----> much easier to read !
3) (4x − 6)^2 = 16x^2 − 24x − 24x + 36 = 16x^2 − 48x + 36
4) (x^2 / y^5)^4 = (x^2)^4 / (y^5)^4 = x^8 / y^20
5) (3x + 5y)(4x − 3y) = 12x^2 − 9xy + 20xy − 15y^2 = 12x^2 + 11xy − 15y^2
6) (3x^3y^4z^4)(2x^3y^4z^2) = (3*2) x^(3+3) y^(4+4) z^(4+2) = 6 x^6 y^8 z^6
7) 5x + 3x^4 − 7x^3 ----> Fourth degree trinomial
8) (5x^3 − 5x − 8) + (2x^3 + 4x + 2) = 7x^3 − x − 6
9) (x − 1) + (2x + 5) − (x + 3) = x + 1
10) (−4g^8h^5k^2)0(hk^2)^2 = 0 (anything multiplied by 0 = 0)
or.. (−4g^8h^5k^2)^0(hk^2)^2 = 1 (h^2 (k^2)^2) = h^2 k^4
Last question shows why it is so important to use proper symbols (such
as ^ to indicate exponents). Without such symbols, I could not tell if
the 0 was an actual number and part of multiplication, of if 0 was an
exponent of the expression preceding it.
P.S
Glad to help you have an AWESOME!!! day :)
</span>
So it would be x12 to the 2nd power, kinda hard to explain but i really hope this helped
