4x3+2 = 14/3 —> 56/12
1x4+3 = 7/4 —> 21/12
56-21 is equal to 35/12
Which is 2 and 11/12
Let the second score = x
⇒ second score = x
<span>the first is 14 points more than the second
</span>⇒ first score = x + 14
<span>the sum of the first two is 6 more than twice the third
</span>⇒ third score = 1/2 (x + x + 14 - 6) = x + 4
<span>The sum of a student's three score is 246
</span>⇒ x + (x + 14) + (x + 4) = 246
<span>
Solve x:
</span>x + x + 14 + x + 4 = 246
3x + 18 = 246
3x = 228
x = 76
second score = x = 76
first score = x + 14 = 76 + 14 = 90
Answer: 90
Answer:
za=24
Step-by-step explanation:
supplementary angles add up to equal 180
180=7x+x+30+6
step 1 combine like terms
180=8x+36
step 2 subtract 36 from each side
144=8x
step 3 divide each side by 8
x=18
now we just plug in 18 to x in x+6
18+6=24
Answer:
Only d) is false.
Step-by-step explanation:
Let
be the characteristic polynomial of B.
a) We use the rank-nullity theorem. First, note that 0 is an eigenvalue of algebraic multiplicity 1. The null space of B is equal to the eigenspace generated by 0. The dimension of this space is the geometric multiplicity of 0, which can't exceed the algebraic multiplicity. Then Nul(B)≤1. It can't happen that Nul(B)=0, because eigenspaces have positive dimension, therfore Nul(B)=1 and by the rank-nullity theorem, rank(B)=7-nul(B)=6 (B has size 7, see part e)
b) Remember that
. 0 is a root of p, so we have that
.
c) The matrix T must be a nxn matrix so that the product BTB is well defined. Therefore det(T) is defined and by part c) we have that det(BTB)=det(B)det(T)det(B)=0.
d) det(B)=0 by part c) so B is not invertible.
e) The degree of the characteristic polynomial p is equal to the size of the matrix B. Summing the multiplicities of each root, p has degree 7, therefore the size of B is n=7.