Answer:
2a^4+5a^3-6a^2+19a-20
Step-by-step explanation:
(a^2+3a-4)(2a^2-a+5)
2a^4-a^3+5a^2+6a^3-3a^2+15a-8a^2+4a-20
2a^4-a^3+6a^3+5a^2-3a^2-8a^2+15a+4a-20
2a^4+5a^3+2a^2-8a^2+19a-20
2a^4+5a^3-6a^2+19a-20
If SU bisects TSV, then TSU = USV
4y + 11 = 6y + 5
6y - 4y = 11 - 5 = 6
y = 6/2 = 3
Therefore, m<TSU = 4(3) + 11 = 12 + 11 = 23
I think it is B because it has a continuous line soy it couldn’t be a. It has negative numbers so it couldn’t be D. And it has negative numbers so couldn’t be C
9514 1404 393
Answer:
x = -9
Step-by-step explanation:
Segment NL is twice the length of midsegment WV.
2WV = NL
2(x+15) = x+21
2x +30 = x +21 . . . . simplify
x = -9 . . . . . . . . . . . . add -30-x
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<em>Additional comment</em>
This value of x means the other segments are ...
MN = 12
WV = 6
NL = 12
First you have to convert 1/10 into a decimal:
1/10= 0.1
Multiply 0.1 by 200:
0.1x200=20
So, 20 is 1/10 of 200.
