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3 years ago

AB is included between--------

Mathematics

2 answers:

Aleksandr [31]3 years ago

7 0

Answer: $\overline{AB}$ included between $\angle{A}$ and $\angle{B}$

Step-by-step explanation:

The included side means the side between two angles. It usually found in triangles and polygons.

In the given figure there is a triangle ABC having three angles and three sides.

By the definition of included side, we have

$\overline{AB}$ is a side included between $\angle{A}$ and $\angle{B}$

$\overline{BC}$ is a side included between $\angle{B}$ and $\angle{C}$

$\overline{CA}$ is a side included between $\angle{C}$ and $\angle{A}$

Hence, $\overline{AB}$ included between $\angle{A}$ and $\angle{B}$

nirvana33 [79]3 years ago

5 0

Answer:

B) ∠A and ∠B

Step-by-step explanation:

AB is the line segment formed with endpoints at A and B. This means it lies between the angle with vertex at A, ∠A, and the angle with vertex at B, ∠B.

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3 years ago

The expression (X + 3)(x + 2) is the product of two binomials. Which expression is also a product of binomials?

kow [346]

Answer:

Step-by-step explanation:

Binomials are expressions containing the sum (or difference) of two terms.

A. is not the answer because pq and qp are <u>products</u>.

C. is not the answer because 3x is not a sum/difference.

D. is not the answer because it is one binomial, not two muliplied together.

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3 years ago

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ololo11 [35]

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Step-by-step explanation:

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2 years ago

Suppose that X is a random variable with mean 30 and standard deviation 4. Also suppose that Y is a random variable with mean 50

Vladimir79 [104]

Answer:

a) Var[z] = 1600

D[z] = 40

b) Var[z] = 2304

D[z] = 48

c) Var[z] = 80

D[z] = 8.94

d) Var[z] = 80

D[z] = 8.94

e) Var[z] = 320

D[z] = 17.88

Step-by-step explanation:

In general

V([x+y] = V[x] + V[y] +2Cov[xy]

how in this problem Cov[XY] = 0, then

V[x+y] = V[x] + V[y]

Also we must use this properti of the variance

V[ax+b] = $a^{2}$ V[x]

and remember that

standard desviation = $\sqrt{Var[x]}$

a) z = 35-10x

Var[z] = $10^{2}$ Var[x] = 100*16 = 1600

D[z] = $\sqrt{1600}$ = 40

b) z = 12x -5

Var[z] = $12^{2}$ Var[x] = 144*16 = 2304

D[z] = $\sqrt{2304}$ = 48

c) z = x + y

Var[z] = Var[x+y] = Var[x] + Var[y] = 16 + 64 = 80

D[z] = $\sqrt{80}$ = 8.94

d) z = x - y

Var[z] = Var[x-y] = Var[x] + Var[y] = 16 + 64 = 80

D[z] = $\sqrt{80}$ = 8.94

e) z = -2x + 2y

Var[z] = 4Var[x] + 4Var[y] = 4*16 + 4*64 = 320

D[z] = $\sqrt{320}$ = 17.88

7 0

3 years ago

Which of the following functions are homomorphisms?

Vikentia [17]

Part A:

Given $f:Z \rightarrow Z,$ defined by $f(x)=-x$

$f(x+y)=-(x+y)=-x-y \\ \\ f(x)+f(y)=-x+(-y)=-x-y$

but

$f(xy)=-xy \\ \\ f(x)\cdot f(y)=-x\cdot-y=xy$

Since, f(xy) ≠ f(x)f(y)

Therefore, the function is not a homomorphism.

Part B:

Given $f:Z_2 \rightarrow Z_2,$ defined by $f(x)=-x$

Note that in $Z_2$ , -1 = 1 and f(0) = 0 and f(1) = -1 = 1, so we can also use the formular $f(x)=x$

$f(x+y)=x+y \\ \\ f(x)+f(y)=x+y$

and

$f(xy)=xy \\ \\ f(x)\cdot f(y)=xy$

Therefore, the function is a homomorphism.

Part C:

Given $g:Q\rightarrow Q$ , defined by $g(x)= \frac{1}{x^2+1}$

$g(x+y)= \frac{1}{(x+y)^2+1} = \frac{1}{x^2+2xy+y^2+1} \\ \\ g(x)+g(y)= \frac{1}{x^2+1} + \frac{1}{y^2+1} = \frac{y^2+1+x^2+1}{(x^2+1)(y^2+1)} = \frac{x^2+y^2+2}{x^2y^2+x^2+y^2+1}$

Since, f(x+y) ≠ f(x) + f(y), therefore, the function is not a homomorphism.

Part D:

Given $h:R\rightarrow M(R)$ , defined by $h(a)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)$

$h(a+b)= \left(\begin{array}{cc}-(a+b)&0\\a+b&0\end{array}\right)= \left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right) \\ \\ h(a)+h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)+ \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)=\left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right)$

but

$h(ab)= \left(\begin{array}{cc}-ab&0\\ab&0\end{array}\right) \\ \\ h(a)\cdot h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)\cdot \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)= \left(\begin{array}{cc}ab&0\\-ab&0\end{array}\right)$

Since, h(ab) ≠ h(a)h(b), therefore, the funtion is not a homomorphism.

Part E:

Given $f:Z_{12}\rightarrow Z_4$ , defined by $\left([x_{12}]\right)=[x_4]$ , where $[u_n]$ denotes the lass of the integer $u$ in $Z_n$ .

Then, for any $[a_{12}],[b_{12}]\in Z_{12}$ , we have

$f\left([a_{12}]+[b_{12}]\right)=f\left([a+b]_{12}\right) \\ \\ =[a+b]_4=[a]_4+[b]_4=f\left([a]_{12}\right)+f\left([b]_{12}\right)$

and

$f\left([a_{12}][b_{12}]\right)=f\left([ab]_{12}\right) \\ \\ =[ab]_4=[a]_4[b]_4=f\left([a]_{12}\right)f\left([b]_{12}\right)$

Therefore, the function is a homomorphism.

7 0

3 years ago