What is angle DCE? A. 34 B. 85 C. 102 D. 98
2 answers:
The red line is 180 degrees. So the arcs on top of the circle all equal up to 180 degrees. the number 72 doesn't really matter cause you're trying to find out the angle of DCE which is on the other side of the circle. So you gotta find the arc length of DE which is 68 because 180-112=68. So to find the angle, you have to divide the opposite arc length by two, which is 34.
ANSWER: A. 34. you're welcomee!!
ANSWER: A. 34. you're welcomee!!
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Answers:
- false
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- True
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Explanation:
Problem 1
This is false because the A and B should swap places. It should be .
The short proof is to multiply AB with its inverse and we get:
The fact we get the identity matrix proves that we have the proper order at this point. The swap happens so that B matches up its corresponding inverse and the two cancel each other out.
Keep in mind matrix multiplication is <u>not</u> commutative. So AB is not the same as BA.
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Problem 2
This statement is true if and only if AB = BA
(A+B)^2 = (A+B)(A+B)
(A+B)^2 = A(A+B) + B(A+B)
(A+B)^2 = A^2 + AB + BA + B^2
(A+B)^2 = A^2 + 2AB + B^2 ... only works if AB = BA
However, in most general settings, matrix multiplication is <u>not</u> commutative. The order is important when multiplying most two matrices. Only for special circumstances is when AB = BA going to happen. In general, AB = BA is false which is why statement two breaks down and is false in general.
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Problem 3
This statement is true.
If A and B are invertible, then so is AB.
This is because both and
are known to exist (otherwise A and B wouldn't be invertible) and we can use the rule mentioned in problem 1. Make sure to swap the terms of course.
Or you can use a determinant argument to prove the claim
det(A*B) = det(A)*det(B)
Since A and B are invertible, their determinants det(A) and det(B) are nonzero which makes the right hand side nonzero. Therefore det(A*B) is nonzero and AB has an inverse.
So if we have two invertible matrices, then their product is also invertible. This idea can be scaled up to include things like A^4*B^3 being also invertible.
If you wanted, you can carefully go through it like this:
- If A and B are invertible, then so is AB
- If A and AB are invertible, then so is A*AB = A^2B
- If A and A^2B are invertible, then so is A*A^2B = A^3B
and so on until you build up to A^4*B^3. Therefore, we can conclude that A^m*B^n is also invertible. Be careful about the order of multiplying the matrices. Something like A*AB is different from AB*A, the first of which is useful while the second is not.
So this is why statement 3 is true.
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Problem 4
This is false. Possibly a quick counter-example is to consider these two matrices
both of which are invertible since their determinant is nonzero (recall the determinant of a diagonal matrix is simply the product along the diagonal entries). So it's not too hard to show that the determinant of each is 1, and each matrix shown is invertible.
However, adding those two mentioned matrices gets us the 2x2 zero matrix, which is a matrix of nothing but zeros. Clearly the zero matrix has determinant zero and is therefore not invertible.
There are some cases when A+B may be invertible, but it's not true in general.
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Problem 5
This is true because each A pairs up with an to cancel out (similar what happened with problem 1). For more info, check out the concept of diagonalization.
Answer:
b =
Step-by-step explanation:
To clear the radicals square both sides
=
( cross- multiply )
(6 - a)(8 + b) = b - 7 ← expand factors on left using FOIL
48 + 6b - 8a - ab = b - 7 ( subtract 48 from both sides )
6b - 8a - ab = b - 55 ( add 8a to both sides )
6b - ab = b - 55 + 8a ( subtract b from both sides )
6b - ab - b = 8a - 55, that is
5b - ab = 8a - 55 ← factor out b from each term on the left side )
b(5 - a) = 8a - 55 ← divide both sides by (5 - a)
b =