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11111nata11111 [884]
3 years ago
14

2^8+6(7-1)+4 does this equal 56

Mathematics
1 answer:
sukhopar [10]3 years ago
7 0

Answer:

nope

Step-by-step explanation:

it acutally equals to 292 I am sure!

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An oarsman can row his boat 3 mph in still water. He sets out on the Illinois River, which flows at 5 mph. We are interested in
Artyom0805 [142]

Answer:

The speed from the observer is 8 mph.

Step-by-step explanation:

When the boat is going downstream with full force, its speed will be the one the oarsman can row on still water added by the flow of the river. In this case he can row the boat at speed of 3 mph, while the stream is 5 mph, so the speed from the observer standpoint is:

speed = 3 + 5\\speed = 8 \text{ mph}

The speed from the observer is 8 mph.

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3 years ago
Solve the equation for x:<br> 4-3/2x= 16<br> MIN
Vera_Pavlovna [14]

Answer: x= -8

Step-by-step explanation:

7 0
3 years ago
How many dollars are in Fifty pennies
andrew11 [14]
1/2. 50 pennies equals $0.50
4 0
3 years ago
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A paper clip is dropped from the top of a 144‐ft tower, with an initial velocity of 16 ft/sec. Its position function is s(t) = −
hoa [83]
To calculate the velocity, we use the given expression above which is <span>s(t) = −16t^2 + 144. First, we calculate the time it takes to reach the ground. Then, differentiate the expression and substitute time to the differentiated expression.

</span>s(t) = −16t^2 + 144
0 = -16t^2 + 144
t = 3

s'(t) = v = -32t
v = -32(3)
v = -96 

Note: negative sign signifies that the object is going down
5 0
2 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
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