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3 years ago

What is the area of a sheet of binder paper? (Binder paper is 8 1/2 inches by 11 inches.)

1 answer:

3 0

Area of a rectangle is the product of length times width. Thus, here we need the product of 8 1/2 inches by 11 inches.

One way of doing this follows: 8.5
8.5
----------
93.5 square inches

Rounding up, this comes out to 94 square inches.

Alternatively: 17 187
---- * 11 = ------- = 93.5 square inches; rounded, that's 94.
2 2

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Mandisa draws a rectangle to represent the area of her yard. The area can be represented by 10x2 – 13x – 14.What expressions can

Y_Kistochka [10]

Answer: (x–2) and (10x+7)

Step-by-step explanation:

10x^2 – 13x – 14=

10x^2 – 20x + 7x – 14=

10x(x–2) + 7(x–2)=(x–2)(10x+7) ==> 3rd option

4 0

2 years ago

The difference between eight times a number and six more than three times the number. Please help find the equation and simplify

Rainbow [258]

Answer: x>1.2

Step-by-step explanation:

Let a number is x.

Hence,

$8x-6 > 3x\\\\8x-6-3x > 3x-3x\\\\5x-6 > 0\\\\5x-6+6 > 0+6\\\\5x > 6\\$

Divide both parts of the equation by 5:

$x > 1.2$

8 0

1 year ago

Which of the following is equivalent to √-90?<br> A. -3i√10<br> B. 3i√10<br> C. 9i√10<br> D. 9√-10

erastova [34]

The answer to this question is B. I hope this helps

7 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

3 years ago

Which equation represents the function graphed on the

Pavel [41]

Answer:

Step-by-step explanation:

sorry im late

8 0

2 years ago