Answer:
This statement is false. Exterior angles are the angles created when the sides of the triangle are extended.
Step-by-step explanation:
Exterior angles are the angle between the any reference side and line extended from the adjacent side.
the sum of exterior and interior angle is 180 as it lies on a straight line.
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based on above definition the given statement becomes false.
Hence, correct choice is option third
This statement is false. Exterior angles are the angles created when the sides of the triangle are extended.
Answer:
I really don't know but can u help with mine-_-
Answer:
110 cm^2
Step-by-step explanation:
The first thing that you need to do is find the area of triangle AFE. The area of a triangle is always base*height/2. So in this case, that would be 10*6 divided by 2, which is 30 cm. Next, you will need to know the area of triangle ECB. Using that same formula, you will get 8*10/2, which is 40 cm. Finally, you will need to find the area of the whole rectangle. The area of a rectangle is always the length times the width. In this case, you would have 10*18, which is 180 cm. To get your final answer, you need to subtract the areas of the unshaded area from the whole area. That would be 180-(30+40), which is 110 cm. I hope this helped!
9514 1404 393
Answer:
$7.14
Step-by-step explanation:
Let p, d, q represent the numbers of pennies, dimes, and quarters in the collection, respectively.
p + d + q = 45 . . . . . . . . there are 45 coins in the collection
2p +5 = q . . . . . . . . . . . . 5 more than twice the number of pennies
p + 4 = d . . . . . . . . . . . . . 4 more than the number of pennies
Substituting the last two equations into the first gives ...
p +(p +4) +(2p +5) = 45
4p = 36 . . . . . . . . . . . . . subtract 9
p = 9 . . . . . . . . . . . divide by 4
d = 9 +4 = 13
q = 2(9) +5 = 23
The value of the collection is ...
23(0.25) +13(0.10) +9(0.01) = 5.75 +1.30 +0.09 = 7.14
The coin collection is worth $7.14.
The rational root theorem states that the rational roots of a polynomial can only be in the form p/q, where p divides the constant term, and q divides the leading term.
In your case, both the leading term 5 and the constant term 11 are primes, so their only divisors are 1 and themselves.
So, the only feasible solutions are
For the record, in this case, none of the feasible solutions are actually a root of the polynomial.
