Your answer is B. You multiply 5 by the exponents and 4^0 is just 1, so 4^45•1=4^45. Hope this helped!
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Answer:
A. ∠EAG and ∠BAC
Step-by-step explanation:
Vertical angles share a vertex and have opposite rays for sides.
Rays AE and AB are opposite; rays AG and AC are opposite. These are the only pairs of opposite rays in the figure, so the vertical angles must be constructed from one of the rays of the first pair and one of the rays of the second pair. ∠EAG and ∠BAC are a pair of vertical angles.
The other pair is ∠EAC and ∠BAG.
Answer:
The fourth term is -102----------------------------------------------
Explanation:
The term after the nth term is generated by this rule
which means that we first
Step 1) multiply the nth term (
) by -4
Step 2) Add the result of step 1 to the value 2 to get the next term in the sequence
Let's follow those steps above to generate the first four terms
The first term is
. In short, the first term is 2
The second term is...
So the second term is -6
The third term is...
The third term is 26
Finally, the fourth term is...
The fourth term is -102.
Answer: y = -2/3(x-3)^2 + 0 or y = -2/3(x-3)^2
Step-by-step explanation:
vertex form is y=a(x-h)^2 + k
here we can see the vertex is (3,0) which is (x,y). Or (h,k) in this case.
so to plug that into vertex form, we now have y=a(x-3)^2 + 0. or just y=a(x-3)^2.
now we need to find "a" which is the leading coefficient. to do that we can plug in the (6,-6) for the x and y parts of the above equation. so we'd have
-6=a(6-3)^2. which goes to -6=a(2)^2 which is -6=4a. divide each side by 4 to get a = -2/3. plug this in for a
the final equation would be y = -2/3(x-3)^2 + 0 or y = -2/3(x-3)^2
Answer:
Mean and IQR
Step-by-step explanation:
The measure of centre gives the central or the measure which gives the best mid term of a distribution. Based in the details of the box plot, the median is the value which divides the box in the box plot.
For company A:
Range = 25 to 80 with a median value at 30 ; this means the median does not give a good centre measure of the distribution ad it is very close to the minimum value. This goes for the Company B plot too; with values ranging from (35 to 90) and the median designated at 40.
Hence, the mean will be the best measure of centre rather Than the median in this case.
For the variability, the interquartile range would best suit the distribution. With the lower quartile and upper quartile both having reasonable width to the minimum and maximum value of the distribution.
