Answer:
C. It would be shifted up.
Step-by-step explanation:
We base it off the y-intercept. Since -7 is <em>less</em><em> </em><em>than</em><em> </em>1, and our new function has a positive y-intercept, it would indeed be shifted up.
I am joyous to assist you anytime.
To solve for the time an object takes to hit the ground, the equation
<em>y</em> <em>= y₀ + v₀t + ½gt²</em>, or just <em>0 = y₀ +½gt²</em> since there is no initial velocity and the final y position is 0 (since it is on the ground). Therefore, <em>t = (−2y₀/g)^½.</em>
(-2*60/-9.8)^½ = 3.50 seconds.
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Answer:
I don't have a protractor with me but if you do look where x is and line the little hole on the bottom of the protractor on the crossing point and see what degree the angle is at. Hope I helped :)
Step-by-step explanation:
Answer: First Option
The slope is −0.4. Starting at 3.8, the GPA will decrease by 0.4 for every text sent in class.
Step-by-step explanation:
The equation of a line has the following form
Where m is the slope of the line and b is the intercept with the axis of the y
In this case, the equation is:
Observe that
So if x represents the number of texts sent during classes, this means that 
When x increases by one unit then y decreases -0.4. For example when 
then 
Then the greatest value it can take the GPA is 3.8. Therefore we can say that the function starts at y = 3.8.
The answer is the first option
Clearly, |S| = 50.
Count the multiples of 2 between 1 and 50:
⌊50/2⌋ = ⌊25⌋ = 25
(where ⌊x⌋ denotes the "floor of x", or the largest integer that is smaller than or equal to x; in other words, round <u>down</u> to the nearest integer)
Count the multiples of 3 between 1 and 50:
⌊50/3⌋ ≈ ⌊16.667⌋ = 16
Since LCM(2, 3) = 6, the sets of multiples of 2 and multiples of 3 have some overlap. Count the multiples of 6 between 1 and 50:
⌊50/6⌋ ≈ ⌊8.333⌋ = 8
Then by the inclusion/exclusion principle, we remove from S
25 + 16 - 8 = 33
elements, so that the new set S contains 50 - 33 = 17 elements.
