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3 years ago

Write a word problem using the numbers 506 and 8. Then solve

Mathematics

1 answer:

natka813 [3]3 years ago

3 0

Answer:

Billy is a fly enthusiast! He has a huge glass container filled with 506 flies! One day when Billy went to feed the flies, he accidentally left the screw opened for too long, and 8 flies escaped! How many flies are left in the tank?

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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

a car manufacturing company was able to release 1860 units at the end of january based on records they were able to increase the

saw5 [17]

Answer:

11th month which is November

Step-by-step explanation:

1860 + 45(m - 1) = 2310

1860 + 45m - 45 = 2310

1815 + 45m = 2310

45m = 495

m = 11

8 0

3 years ago

ophia can text 95 words in 10 minutes. At this rate, how many minutes would it take her to text 114 words? Fill out the table of

Usimov [2.4K]

Answer:

12 minuts

Step-by-step explanation

If you divide 95 by ten you get 9.5, and 9.5 to 95 and you get 104.5 and it is 11 minuts. add it again and you are at 114 words and it is 12 minuts.

4 0

2 years ago

(Click on the pictures for the question)

harkovskaia [24]

Answer:

1. -1,1 -2,2 -3,3. 2. when we are digging and when we are planting something. 3. Zero is the center point in the negitives, It tells you when you are about to hit negitives or positives.

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

You have been tasked with filling 4 ounce and 3 ounce bags from a 41 ounce container of candy.

Alchen [17]

Answer:

A. 3 possible combinations

B. 8 4-ounce's bags and 3 3-ounce's bags

C. 2 4-ounce's bags and 11 3-ounce's bags

D. 8 4-ounce's bags and 3 3-ounce's bags

E. All solutions offer the same revenue.

Step-by-step explanation:

You have been tasked with filling 4 ounce and 3 ounce bags from a 41 ounce container of candy. Let x be the number of 4 ounce bags and y be the number of 3 ounce bags. Then

$4x+3y=41.$

A. Find all integer solutions:

When x=0, then 3y=41 - impossible, because 41 is not divisible by 3.

When x=1, then 3y=37 - impossible, because 37 is not divisible by 3.

When x=2, then 3y=33, y=11 - possible.

When x=3, then 3y=29 - impossible, because 29 is not divisible by 3.

When x=4, then 3y=25 - impossible, because 25 is not divisible by 3.

When x=5, then 3y=21, y=7 - possible.

When x=6, then 3y=17 - impossible, because 17 is not divisible by 3.

When x=7, then 3y=13 - impossible, because 13 is not divisible by 3.

When x=8, then 3y=9, y=3 - possible.

When x=9, then 3y=5 - impossible, because 5 is not divisible by 3.

When x=10, then 3y=1 - impossible, because 1 is not divisible by 3.

You get 3 possible combinations.

B. 1. 2 + 11 = 13,

2. 5 + 7 = 12,

3. 8 + 3 = 11.

The minimal number of bags is 11.

C. 1. 2·7+11·5=69 cents

2. 5·7+7·5=70 cents

3. 8·7+3·5=71 cents

The cheapest is 1st solution.

D. 1. 2·6+11·5=67 cents

2. 5·6+7·5=65 cents

3. 8·6+3·5=63 cents

The cheapest is 3rd solution.

E. 1. 2·2+11·1.50=$20.50

2. 5·2+7·1.50=$20.50

3. 8·2+3·1.50=$20.50

All solutions offer the same revenue.

5 0

2 years ago