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3 years ago

Marsha used 6 cups of flour to make 4 trays of

Mathematics

1 answer:

PolarNik [594]3 years ago

6 0

Answer:

Step-by-step explanation:

4/6 = 1.5, 1.5(6)= 9

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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

Solve for x: 3√5x² +25x-10√√5 = 0

bogdanovich [222]

Answer: x = - 2√5, √5/3

Step-by-step explantion

3 √ 5 x 2 + 25 x − 10 √ 5 = 0 35x2+25x−105=0

⇒ 3√5x2 + 30x – 5x – 10√5 = 0

⇒ 3√5x(x + 2√5) – 5(x + 2√5) = 0

⇒ (x + 2√5)(3√5x – 5) = 0

x = - 2√5, √5/3

3 0

1 year ago

Read 2 more answers

Write each of the following as a polynomial in descending powers of its variable

nikklg [1K]

1. 7x(x + 3)
    7x(x) + 7(3)
    7x² + 21

2. (7x + 1)(x + 3)
    7x² + 21x + x + 3
    7x² + 22x + 3

3. (x² + x)²
    (x² + x)(x² + x)
   x^4 + x³ + x³ + x²
   x^4 + 2x³ + x²

4 0

3 years ago

What is 8 divided by 108

patriot [66]

The answer is 13.5, hope this helps!

6 0

3 years ago

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If r = 4 and ^a8 = 100, what is the first term of the geometric sequence?

telo118 [61]

The first term of the geometric sequence is 0.0061

Step-by-step explanation: