<span>The number rounded all the way is 3,000. If you round to the hundreds you have to remove the units number which is 9, then you have to increase the tens number to 9, resulting 2,690. Next you have to remove the number 9 and increase the 6 to 7, obtaning 2,700. Finally, you remove the seven and have to incrase the number 2 to 3, to get 3,000. This is three thousdand. </span>
3 is incorrect because 14.7 + 3 = 17.7
The answer of 15 - 14.7 = 0.3
Vertical angles are equal...so set ur angles equal to each other and solve for x
5x + 10 = 7x - 12
12 + 10 = 7x - 5x
22 = 2x
22/2 = x
11 = x <==
To find out end behavior, we always look at the largest exponent (3 in this case) and the sign of the term with the largest exponent (negative in this case). We can rule out the first two choices because since it is a cubic graph (exponent of 3) the graph will be pointing in opposite directions. Then, the negative tells us that as the x values approache infinity, the y values will approach negative infinity. So the correct answer is the last choice.
Hope this helps
Answer:
A and C
Step-by-step explanation:
To determine which events are equal, we explicitly define the elements in each set builder.
For event A
A={1.3}
for event B
B={x|x is a number on a die}
The possible numbers on a die are 1,2,3,4,5 and 6. Hence event B is computed as
B={1,2,3,4,5,6}
for event C
![C=[x|x^{2}-4x+3]\\solving x^{2}-4x+3\\x^{2}-4x+3=0\\x^{2}-3x-x+3=0\\x(x-3)-1(x-3)=0\\x=3 or x=1](https://tex.z-dn.net/?f=C%3D%5Bx%7Cx%5E%7B2%7D-4x%2B3%5D%5C%5Csolving%20%20x%5E%7B2%7D-4x%2B3%5C%5Cx%5E%7B2%7D-4x%2B3%3D0%5C%5Cx%5E%7B2%7D-3x-x%2B3%3D0%5C%5Cx%28x-3%29-1%28x-3%29%3D0%5C%5Cx%3D3%20or%20x%3D1)
Hence the set c is C={1,3}
and for the set D {x| x is the number of heads when six coins re tossed }
In the tossing a six coins it is possible not to have any head and it is possible to have head ranging from 1 to 6
Hence the set D can be expressed as
D={0,1,2,3,4,5,6}
In conclusion, when all the set are compared only set A and set C are equal
