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3 years ago

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:

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