All categories

2 years ago

The decimal 0.3 represents 1/3. What type of number best describes 0.9, which is 3 times 0.3? Explain.

Mathematics

2 answers:

myrzilka [38]2 years ago

8 0

It can be described as 9/10

ss7ja [257]2 years ago

4 0

Answer:

$0.9=\frac{9}{10}$

Step-by-step explanation:

We are given that decimal number 0.3 represents $\frac{1}{3}$

We have to fine the number that describes best 0.9 and 3 times of 0.3

We are given that the number which descries best 0.9 is 3 times of 0.3

$0.3=\frac{1}{3}$

$0.9=\frac{9}{10}$

$0.3=\frac{3}{10}$

Multiply by 3 then we get

$0.3\times 3=\frac{3}{10}\times 3$

$0.9=\frac{9}{10}$

Hence, the number is $\frac{9}{10}$ which describes best 0.9.

You might be interested in

Higher order thinking: Jeff finds some bugs. He finds 10 Fewer grasshoppers than crickets. He finds 5 fewer crickets than Lady b

Stells [14]

C - 10 = g
L - 5 = C

g = 5

then altogether he finds

40 bugs

8 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 an equation in point-slope form for the line that passes through the point with the given slope. point: (4,2) and m=3

jonny [76]

Answer:

y = 3x - 10

Step-by-step explanation:

4 0

3 years ago

I'm a factor of 10 but not a multiple of 5

makkiz [27]

2 it is 2 and thats all it will ever be man