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3 years ago

I need to find the area of these shapes

Mathematics

1 answer:

Wewaii [24]3 years ago

6 0

Answer:

1: 374cm

2: 248mm

3: 69m

I believe this is the right answer but I don't think the labels are correct

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2. Write an expression using fractions to show how to determine the amount that each person will pay. Then calculate each person

meriva

Makaylah: 7371 / 250

Keisha: 4731 / 250

Kelviona: 8871 /250

Janell: 2217 /125

(i dont know the divison part)

3 0

3 years ago

Find the slope from the table below.

lorasvet [3.4K]

Slope for the function is -1/3.5.

6 0

3 years ago

Read 2 more answers

Please help me I need an answer

Mumz [18]

1) 2/4 = 1/4 + 1/4
2) 4/6 = 1/6 + 1/6 + 1/6 + 1/6
3) 3/5 = 1/5 + 1/5 + 1/5
4) 3/3 = 1/3 + 1/3 + 1/3
5) 7/8 = 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8
6) 6/2 = 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2
7) 5/6 = 1/6 + 1/6 + 1/6 + 1/6 + 1/6
8) 9/5 = 1/5 + 1/5 + 1/5 + 1/5 + 1/5 + 1/5 + 1/5 + 1/5 + 1/5
9) 8/3 = 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3
10) 1/2 * 8 = 8/2 = 4

11) Unit fractions are fractions where its numerators are 1 and its denominators are positive integer. example: 1/2 ; 1/3 ; 1/4

12) D.) 5/2 = 5 * 1/2

7 0

3 years ago

Which triangle is a right triangle?

iren [92.7K]

Answer:

A right triangle is always 90 degrees.

Step-by-step explanation:

4 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

4 years ago