D=4q
25q+10d=195
5q+2d=39
sub 4q foro d
5q+2(4q)=39
5q+8q=39
13q=39
divide 13
q=3
d=4q
d=4(3)
d=12
12 dimes
3 quarters
(13x+14)(6x-5)=0
78x²-65x+84x-70=0
78x²+19x-70=0
It was equation.
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Factoring⇒ a×c=78×(-70)=-5460
84×(-65)=-5460
84+(-65)=19 which is b
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78x²+84x-65x-70=0
6x(13x+14)-5(13x+14)=0
(13x+14)(6x-5)=0
13x+14=0 6x-5=0
13x=-14 6x=5
x=-14/13 x=5/6
Two solution X=-14/13;5/6
Is there a picture that would help me so I can help you? Thanks
Answer:
Step-by-step explanation:
T is a linear transformation, hence it is homogeneous (T(cr)=cT(r) for all real c and r∈ℝ³) and additive (T(r+s)=T(r)+T(s), for all r,s∈ℝ³). Apply these properties with r=3u and s=2v to obtain:
We don't have an explicit definition of T, so it's more difficult to compute T(3u+2v) directly without using these properties.
Answer: C. 625 and 81
Step-by-step explanation: A relatively prime pair is a pair in which in both numbers given, the only number that can go into each number is one. In 112 and 36, we know 2 can evenly go into each number because they both end in a positive number (112/2 is 56 and 36/2 is 18.) This means A is incorrect. As for B, 11 can evenly go into each number, leaving you with 25 and 7. B is incorrect. And then there's D. 5 can go into each number giving you 160 and 19. D is incorrect.
C is correct because the only numbers that can go evenly into 81 is 1, 3, 9, 27, and 81. None of these numbers, except one can go into 625 also.
