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Olegator [25]
3 years ago
8

Find the unknown sides and angles of this triangle using the Law of Cosines.

Mathematics
1 answer:
zaharov [31]3 years ago
7 0

Answer:

side c-10.35 Angle a-50.88 Angle B-94.12

Step-by-step explanation:

I found the answers on a khan academy thing. I just took the quiz and got a 100%

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The product-3 and t is greater than 11
PSYCHO15rus [73]

Answer:

i am pretty sure its 8

Step-by-step explanation:

because a negative plus a positive = wichever one is bighger and 8 is bigger than 3 so 11 is positive. 8 +3 = 11

6 0
3 years ago
Helppppppppppppppp plzzzzzzzzzzzzzz
Pavlova-9 [17]

Answer: 15w

Step-by-step explanation:

6 0
3 years ago
Help !! what is the expected value for the binomial distribution below ?
Ierofanga [76]

Answer:

4

Step-by-step explanation:

0*(1/3125)+1*(4/625)+2*(32/625)+3*(128/625)+4*(256/625)+5*(1024/3125)

4 0
3 years ago
Increase £139 by 28%<br> Give your answer rounded to 2 DP
Elis [28]

Answer:

£177.92

Step-by-step explanation:

To increase something by 28% would be to multiply the original value by 1.28:

So to increase £139 by 28%, we do:

£139 × 1.28 = £177.92

So this is our final answer

4 0
2 years ago
Determine if the given mapping phi is a homomorphism on the given groups. If so, identify its kernel and whether or not the mapp
shtirl [24]

Answer:

(a) No. (b)Yes. (c)Yes. (d)Yes.

Step-by-step explanation:

(a) If \phi: G \longrightarrow G is an homomorphism, then it must hold

that b^{-1}a^{-1}=(ab)^{-1}=\phi(ab)=\phi(a)\phi(b)=a^{-1}b^{-1},

but the last statement is true if and only if G is abelian.

(b) Since G is abelian, it holds that

\phi(a)\phi(b)=a^nb^n=(ab)^{n}=\phi(ab)

which tells us that \phi is a homorphism. The kernel of \phi

is the set of elements g in G such that g^{n}=1. However,

\phi is not necessarily 1-1 or onto, if G=\mathbb{Z}_6 and

n=3, we have

kern(\phi)=\{0,2,4\} \quad \text{and} \quad\\\\Im(\phi)=\{0,3\}

(c) If z_1,z_2 \in \mathbb{C}^{\times} remeber that

|z_1 \cdot z_2|=|z_1|\cdot|z_2|, which tells us that \phi is a

homomorphism. In this case

kern(\phi)=\{\quad z\in\mathbb{C} \quad | \quad |z|=1 \}, if we write a

complex number as z=x+iy, then |z|=x^2+y^2, which tells

us that kern(\phi) is the unit circle. Moreover, since

kern(\phi) \neq \{1\} the mapping is not 1-1, also if we take a negative

real number, it is not in the image of \phi, which tells us that

\phi is not surjective.

(d) Remember that e^{ix}=\cos(x)+i\sin(x), using this, it holds that

\phi(x+y)=e^{i(x+y)}=e^{ix}e^{iy}=\phi(x)\phi(x)

which tells us that \phi is a homomorphism. By computing we see

that  kern(\phi)=\{2 \pi n| \quad n \in \mathbb{Z} \} and

Im(\phi) is the unit circle, hence \phi is neither injective nor

surjective.

7 0
3 years ago
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