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2 years ago

Given that the length of the figure below is x + 2, its width is

Mathematics

1 answer:

Phoenix [80]2 years ago

4 0

Answer:

hear is your answer in attachment please give me some thanks

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2 * 10^-4 - 1 * 10^-5 Show your work.

gladu [14]

2 * 10^-4 - 1 * 10^-5
= 10^-4 ( 2 - 1 * 10^-1)
= 10^-4 ( 2 - 0.1)
= 10^-4 (1.9)
= 1.9 * 10^-4

hope it helps

7 0

3 years ago

The term coefficient refers to

Dmitry_Shevchenko [17]

Answer:

It refers to a numerical or constant quantity placed before and multiplying the variable in an algebraic expression.

5 0

2 years ago

4 math problem

Aneli [31]

Answer:

Step-by-step explanation:

1)y=-6x-14 ---------(i)

-x-3y=-9 ---------- (ii)

Substitution method:

Substitute y value in equ (i)

-x -3*(-6x-14) = -9

-x - 3*-6x -3*(-14)= -9

-x + 18x + 42 = -9

17x = -9 -42

17x = -51

x = -51/17

x = -3

Substitute x value in equation (i)

y = -6*-3 -14

y =18-14

y = 4

6 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

Write the product in its simplest form -3y.3y4 Enter the correct answer.

Sauron [17]

Answer:

$\large\boxed{-9y^5}$

Step-by-step explanation:

$-3y\cdot3y^4=(-3\cdot3)(y\cdot y^4)\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=-9y^{1+4}=-9y^5$

5 0

3 years ago