Cost of the current plan: $175
Number of devices: x
Cost of the new plan: $94+($4.50/device)x
Cost of the new plan is less than the current plan:
$94+($4.50/device)x<$175
Solving for x:
$94+($4.50/device)x-$94<$175-$94
($4.50/device)x<$81
(device/$4.50)($4.50/device)x<(device/$4.50)($81)
x<18 devices
Please, see the attached file.
Thanks.
Answer:
Decagon has ten sides and angles.
Answer:
between 25 and 30$_-------------
Answer:
5536 calculators
Step-by-step explanation:
We integrate the function dx/dt to obtain the number of new calculators between beginning of the 3rd week and end of week 4. Note that beginning of 3rd week is the same as end of 2nd week. So,
=
Let u = t + 12, then
= 1. So, du = dt. We also change the limits of our integration. So, when t = 2, u = 2 + 12 = 14 and when t = 4, u = 4 + 12 = 16
Then
= ∫₁₄¹⁶
₁₄¹⁶ = ![5000[16 + \frac{100}{16} - (14 + \frac{100}{14} )] = 5000 [16 - 14 + \frac{100}{16} - \frac{100}{14} ] = 5000 [2 + \frac{100}{16} - \frac{100}{14} ] = 5535.7](https://tex.z-dn.net/?f=5000%5B16%20%2B%20%5Cfrac%7B100%7D%7B16%7D%20-%20%2814%20%2B%20%5Cfrac%7B100%7D%7B14%7D%20%29%5D%20%3D%205000%20%5B16%20-%2014%20%2B%20%5Cfrac%7B100%7D%7B16%7D%20-%20%5Cfrac%7B100%7D%7B14%7D%20%20%5D%20%3D%205000%20%5B2%20%2B%20%5Cfrac%7B100%7D%7B16%7D%20-%20%5Cfrac%7B100%7D%7B14%7D%20%20%5D%20%3D%205535.7)
≈ 5536 calculators
He is wrong because if a + b > c it doesn't follow that a^2 + b^2 > c^2
(a + b^2 will always be greater than c^2 but a^2 + b^2 will not always be.
.
An example:
Right angled triangle with sides 3, 4 and 5:-
3 + 4 > 5 but 3^2 + 4^2 = 5^2