Let the first angle be A and the second be B.
Complementary angles add up to 90° so:
A + B = 90
We are also told:
A = 6 + 3B; substituting this into the first equation:
6 + 3B + B = 90
B = 21°
A = 6 + 3(21)
A = 69°
Answer: welpies
Step-by-step explanation:
I need points so no help.
11. Factoring and solving equations
- A. Factor-
1. Factor 3x2 + 6x if possible.
Look for monomial (single-term) factors first; 3 is a factor of both 3x2
and 6x and so is x . Factor them out to get
3x2 + 6x = 3(x2 + 2x1 = 3x(x+ 2) .
2. Factor x2 + x - 6 if possible.
Here we have no common monomial factors. To get the x2 term
we'll have the form (x +-)(x +-) . Since
(x+A)(x+B) = x2 + (A+B)x + AB ,
we need two numbers A and B whose sum is 1 and whose product is
-6 . Integer possibilities that will give a product of -6 are
-6 and 1, 6 and -1, -3 and 2, 3 and -2.
The only pair whose sum is 1 is (3 and -2) , so the factorization is
x2 + x - 6 = (x+3)(x-2) .
3. Factor 4x2 - 3x - 10 if possible.
Because of the 4x2 term the factored form wli be either
(4x+A)(x +B) or (2x+A)(2x+B) . Because of the -10 the integer possibilities
for the pair A, B are
10 and -1 , -10 and 1 , 5 and -2 . -5 and 2 , plus each of
these in reversed order.
Check the various possibilities by trial and error. It may help to write
out the expansions
(4x + A)(x+ B) = 4x2 + (4B+A)x + A8
1 trying to get -3 here
(2x+A)(2x+B) = 4x2 + (2B+ 2A)x + AB
Trial and error gives the factorization 4x2 - 3x - 10 - (4x+5)(x- 2) .
4. Difference of two squares. Since (A + B)(A - B) = - B~ , any
expression of the form A' - B' can be factored. Note that A and B
might be anything at all.
Examples: 9x2 - 16 = (3x1' - 4' = (3x +4)(3x - 4)
x2 - 29 = x2 - (my)* = (x+ JTy)(x- my)
This equation simplified would be
x= -5/18
<h3>
Answer: 40</h3>
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Explanation:
JQ is longer than QN. We can see this visually, but the rule for something like this is the segment from the vertex to the centroid is longer compared to the segment that spans from the centroid to the midpoint.
See the diagram below.
The ratio of these two lengths is 2:1, meaning that JQ is twice as long compared to QN. This is one property of the segments that form when we construct the centroid (recall that the centroid is the intersection of the medians)
We know that JN = 60
Let x = JQ and y = QN
The ratio of x to y is x/y and this is 2/1
x/y = 2/1
1*x = y*2
x = 2y
Now use the segment addition postulate
JQ + QN = JN
x + y = 60
2y + y = 60
3y = 60
y = 60/3
y = 20
QN = 20
JQ = 2*y = 2*QN = 2*20 = 40
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We have
JQ = 40 and QN = 20
We see that JQ is twice as larger as QN and that JQ + QN is equal to 60.
