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3 years ago

HELP I’m so confused

Mathematics

1 answer:

Mkey [24]3 years ago

5 0

A=7.07107
b=8.48528
c=6.7082

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39x38 multiply using partial prodcuts

yKpoI14uk [10]

39 x 10
= 390

39 x 40 (390 x 4)
= 1,560

39 x 2
= 78

1,560 - 78
= 1,482

ANSWER: 1,482

4 0

3 years ago

What is the circumference of a circle with a radius of 1.4 m?

malfutka [58]

Ok now that i understand the question better i say it is D

4 0

3 years ago

Read 2 more answers

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

4 years ago

The mean amount spent by a family of four on food per month is $500 with a standard deviation of $75. Assuming that the food cos

JulijaS [17]

Answer:

The probability that a family spends less than $410 per month

P( X < 410) = 0.1151

Step-by-step explanation:

Step(i):-

Given mean of the population = 500

Given standard deviation of the Population = 75

Let 'X' be the variable in normal distribution

$Z = \frac{x-mean}{S.D}$

Given X = $410

$Z = \frac{410-500}{75} = - 1.2$

Step(ii):-

The probability that a family spends less than $410 per month

P( X < 410) = P( Z < - 1.2 )

= 0.5 - A( -1.2)

= 0.5 - A(1.2)

= 0.5 - 0.3849 ( ∵from normal table)

= 0.1151

Final answer:-

The probability that a family spends less than $410 per month

P( X < 410) = 0.1151

6 0

3 years ago

Help me solve this please ill give good rating...

blagie [28]

Answer: For the first part, her total savings would be $88, and for the second part, (let's pretend s = the total savings) the equation would be s = 40 + w x 6.

Step-by-step explanation: As for the first part, we would need to multiply her money per week times the amount of weeks she works. We can do this by simply multiplying the amount she makes (6) by the amount of weeks she works (8), resulting in 48, but we still have to add that number to the amount she already has, or 40, making $88 in total, as for the second part, this does a great job of explaining the reasoning behind that as well. Hope this answered your question!

8 0

4 years ago