Given the table showing the distance Randy drove on one day of her vacation as follows:
![\begin{tabular} {|c|c|c|c|c|c|} Time (h)&1&2&3&4&5\\[1ex] Distance (mi)&55&110&165&220&275 \end{tabular}](https://tex.z-dn.net/?f=%5Cbegin%7Btabular%7D%0A%7B%7Cc%7Cc%7Cc%7Cc%7Cc%7Cc%7C%7D%0ATime%20%28h%29%261%262%263%264%265%5C%5C%5B1ex%5D%0ADistance%20%28mi%29%2655%26110%26165%26220%26275%0A%5Cend%7Btabular%7D)
The rate at which she travels is given by
If Randy has driven for one more hour at the same rate, the number of hours she must have droven is 6 hrs and the total distance is given by
distance = 55 x 6 = 330 miles.
Answer: 58unit²
Step-by-step explanation:
We will take 1 square as 1 unit
We will count the number of squares in in this figure.
Total squares = 58
Area = 58unit²
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Answer:
See explanation and hopefully it answers your question.
Basically because the expression has a hole at x=3.
Step-by-step explanation:
Let h(x)=( x^2-k ) / ( hx-15 )
This function, h, has a hole in the curve at hx-15=0 if it also makes the numerator 0 for the same x value.
Solving for x in that equation:
Adding 15 on both sides:
hx=15
Dividing both sides by h:
x=15/h
For it be a hole, you also must have the numerator is zero at x=15/h.
x^2-k=0 at x=15/h gives:
(15/h)^2-k=0
225/h^2-k=0
k=225/h^2
So if we wanted to evaluate the following limit:
Lim x->15/h ( x^2-k ) / ( hx-15 )
Or
Lim x->15/h ( x^2-(225/h^2) ) / ( hx-15 ) you couldn't use direct substitution because of the hole at x=15/h.
We were ask to evaluate
Lim x->3 ( x^2-k ) / ( hx-15 )
Comparing the two limits h=5 and k=225/h^2=225/25=9.
I think you mean y=ab^x.
First, plug in the values of x=2 and y=24 for the first equation, and x=3 and y=48 for the second one:
24 = ab^2
48 = ab^3
divide equation 2 by equation 1:
2 = b^1
or, more simply:
2 = b
now plug b into one of the equations to get "a":
24 = a*2^2
24 = a*4
6 = a
So, the answer is y = 6(2)^x
(I think you need to specify the power of x in your answers)
Hello, your answer would be (-5, 5)
