This result is actually true for any exterior angle. The exterior angle of a triangle is equal to the sum of the two remote angles, and above is a short proof of it.
Answer:
B
Step-by-step explanation:
this is the answer to this question
Answer:
6°F
58°F
Step-by-step explanation:
From the data Given :
Temperature rise in Buffalo by noon time = 14°F
The initial temperature at Buffalo = - 8°F
Rise in temperature = 14°F
Therefore, temperature at noon = (-8 + 14)°F = 6°F
B.)
Temperature at 6am in St. Louis = 32°F
Temperature at 6am in Juneau = - 26°F
Temperature difference :
(32 - (-26))°F
32 + 26
= 58°F
Temperature in St. Louis is 58°F higher
Answer:
c) -x^3 + x^2 - 1
Step-by-step explanation:
Given: u (x) = x^5 - x^4 +x^2 and v(x) = -x^2
(u/v)(x) = u(x)/v(x)
Now plug in the given functions in the above formula, we get
= (x^5 - x^4 + x^2) / -x^2
We can factorize the numerator.
In x^5 - x^4 + x^2. the common factor is x^2, so we can take it out and write the remaining terms in the parenthesis.
= x^2 (x^3 - x^2 + 1) / - x^2
Now we gave x^2 both in the numerator and in the denominator, we can cancel it out.
(u/v)(x) = (x^3 - x^2 + 1) / -1
When we dividing the numerator by -1, we get
(u/v)(x) = -x^3 + x^2 - 1
Answer: c) -x^3 + x^2 - 1
Hope you will understand the concept.
Thank you.
Not is posible because no exist a igualty
