Answer:
10
Step-by-step explanation:
hoh(x) = h(h(x)) = 6 - h(x) = 6 - (6-x) = 6 - 6 + x = x
then
hoh(x) = x
then
hoh(10) = 10
If the table is:
x | y
------
1 | 76
------
2 | 92
---------
3 | 108
-----------
4 | 124
-------------------------------------------------------------------------------------------------------------------
Find the change inbetween each amount per week
92 - 76 = 16
108 - 92 = 16
etc.
This means that her initial deposit should be subtracting 16 from her first deposit.
76 is the amount first deposited
76 - 16 = 60
A) $60 should be your answer
hope this helps
Answer:
13 (c)
Step-by-step explanation:
Graph the equation 7000(1-0.19)^x and then the inequality y<500, the point of intersection should be 12.524, so the answer will be rounded up to 13 i think
the ratio in which 42 should be divided is 1:2:3
the sum of the parts of the ratio is - 1 + 2 + 3 = 6
this means that there's a sum of 6 parts
so we need to find how much 1 part is equivalent to
if 6 parts are equivalent to 42
then 1 part is equivalent to - 42/6 = 7
so the ratio should be 1:2:3
1 part - 7
2 parts - 7 x 2 = 14
3 parts - 7 x 3 = 21
therefore 42 divided into 1:2:3 ratio is as follows
7 : 14 : 12
For the answer to the question above, <span>f x is the number of days she works, she'll earn $90x </span>
<span>After buying the laptop, she'll have $90x - $700 left over, which will pay for ($90x - $700) / $150 days of travel. So we have y = ($90x - $700) / $150 = (9x - 70) / 15 = 0.6x - (14/3) </span>
<span>Note that y can't be negative. Also, if y = 0, then Emma doesn't get to travel at all, so we should avoid that. So we have: </span>
<span>0.6x - (14/3) > 0 </span>
<span>0.6x > 14/3 </span>
<span>x > (14/3) / 0.6 </span>
<span>x > 70/9 </span>
<span>The question says that x can be up to 40, so the domain is 70/9 < x <= 40 </span>
<span>That's approximately 7.777... < x <= 40 </span>
<span>Multiply those numbers by 0.6 and then subtract 700 to get the range: </span>
<span>0 < y <= 58/3 </span>
<span>That's approximately 0 < y <= 19.333</span>
