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Aloiza [94]
3 years ago
10

Anyone who can help please do so but if you dont know the answer don't comment

Mathematics
1 answer:
KonstantinChe [14]3 years ago
8 0
Tough one give me a couple min.
You might be interested in
Please help me fast
Sveta_85 [38]

Answer:

1.187991

Step-by-step explanation:

sin^2(225)−(cos(330))(cos(240))

=(−0.930095)2−(cos(330))(cos(240))

=0.865076−(cos(330))(cos(240))

=0.865076−−0.991199(cos(240))

=0.865076−(−0.991199)(0.325781)

=0.865076−(−0.322914)

=1.187991

8 0
2 years ago
Please solve the questions
Wewaii [24]

Answer:

Step-by-step explanation:

For a

2 + x = 15

x = 15 -2

x = 13

For B

x - 12 = 14

x = 12 + 14

x = 26

For C

3x = 12

x = 12/3

x = 4

Hope it helps:)

5 0
3 years ago
1. Let a; b; c; d; n belong to Z with n > 0. Suppose a congruent b (mod n) and c congruent d (mod n). Use the definition
lukranit [14]

Answer:

Proofs are in the explantion.

Step-by-step explanation:

We are given the following:

1) a \equi b (mod n) \rightarrow a-b=kn for integer k.

1) c \equi  d (mod n) \rightarrow c-d=mn for integer m.

a)

Proof:

We want to show a+c \equiv b+d (mod n).

So we have the two equations:

a-b=kn and c-d=mn and we want to show for some integer r that we have

(a+c)-(b+d)=rn. If we do that we would have shown that a+c \equiv b+d (mod n).

kn+mn   =  (a-b)+(c-d)

(k+m)n   =   a-b+ c-d

(k+m)n   =   (a+c)+(-b-d)

(k+m)n  =    (a+c)-(b+d)

k+m is is just an integer

So we found integer r such that (a+c)-(b+d)=rn.

Therefore, a+c \equiv b+d (mod n).

//

b) Proof:

We want to show ac \equiv bd (mod n).

So we have the two equations:

a-b=kn and c-d=mn and we want to show for some integer r that we have

(ac)-(bd)=tn. If we do that we would have shown that ac \equiv bd (mod n).

If a-b=kn, then a=b+kn.

If c-d=mn, then c=d+mn.

ac-bd  =  (b+kn)(d+mn)-bd

          =    bd+bmn+dkn+kmn^2-bd

          =           bmn+dkn+kmn^2

          =            n(bm+dk+kmn)

So the integer t such that (ac)-(bd)=tn is bm+dk+kmn.  

Therefore, ac \equiv bd (mod n).

//

3 0
3 years ago
A ball is thrown into the air from a height of 9cm above the ground, and its height is given by S=128+32t-16t^2, where t is the
OverLord2011 [107]
-16t² + 32t + 128 = 0
-16t² + 64t - 32t + 128 = 0
-16t(t - 4) - 32(t - 4)
(-16t - 32)(t - 4) = 0
-16t -32 = 0      t - 4 = 0
-16t = 32          t = 4
t = -2

t = -2 and t = 4 are the values that makes S = 0.
Yes, because if you apply either of  these values alone, S will be 0.
6 0
3 years ago
I need help with this problem
sertanlavr [38]

Answer:

Step-by-step explanation:

ill help

6 0
3 years ago
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