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svp [43]
3 years ago
13

My dad disconnected my wifi off my pc with our wifis own app does anyone know how to reconnect my pc to the wifi?

Mathematics
1 answer:
Inga [223]3 years ago
4 0

fijate tu modem a ver si se soluciona si no al tecnico

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What's -18÷(1-4+-2-4) explain pls need help​
Mrac [35]

Answer:

the answer is 2.

Step-by-step explanation:

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Is this relation a function? EXPLAIN WHY OR WHY NOT<br> {(-3,2), (-4, 4), (-3, 3), (4,4)}
denis-greek [22]
No it is not a function, because a function cannot have more than one output per input. When x=-3, there are two solutions shown: (-3,2) and (-3,3), therefore it cannot be a function.
3 0
3 years ago
What is the ratio for cos B
nataly862011 [7]

Answer:

cos B = 8/17

Step-by-step explanation:

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6 0
3 years ago
What is the least common multiple of 14, 20, and 15?
Komok [63]

14, 15, 20 | 2

7, 15, 10 | 2

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1, 1, 1

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Hope it helped,

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7 0
3 years ago
Read 2 more answers
The volume of a right circular cone with radius r and height h is V = pir^2h/3.
Scorpion4ik [409]

The question is incomplete. The complete question is :

The volume of a right circular cone with radius r and height h is V = pir^2h/3. a. Approximate the change in the volume of the cone when the radius changes from r = 5.9 to r = 6.8 and the height changes from h = 4.00 to h = 3.96.

b. Approximate the change in the volume of the cone when the radius changes from r = 6.47 to r = 6.45 and the height changes from h = 10.0 to h = 9.92.

a. The approximate change in volume is dV = _______. (Type an integer or decimal rounded to two decimal places as needed.)

b. The approximate change in volume is dV = ___________ (Type an integer or decimal rounded to two decimal places as needed.)

Solution :

Given :

The volume of the right circular cone with a radius r and height h is

$V=\frac{1}{3} \pi r^2 h$

$dV = d\left(\frac{1}{3} \pi r^2 h\right)$

$dV = \frac{1}{3} \pi h \times d(r^2)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

a). The radius is changed from r = 5.9 to r = 6.8 and the height is changed from h = 4 to h = 3.96

So, r = 5.9  and dr = 6.8 - 5.9 = 0.9

     h = 4  and dh = 3.96 - 4 = -0.04

Now, $dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi (5.9)(4)(0.9)+\frac{1}{3} \pi (5.9)^2 (-0.04)$

$dV=44.484951 - 1.458117$

$dV=43.03$

Therefore, the approximate change in volume is dV = 43.03 cubic units.

b).  The radius is changed from r = 6.47 to r = 6.45 and the height is changed from h = 10 to h = 9.92

So, r = 6.47  and dr = 6.45 - 6.47 = -0.02

     h = 10  and dh = 9.92 - 10 = -0.08

Now, $dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi (6.47)(10)(-0.02)+\frac{1}{3} \pi (6.47)^2 (-0.08)$

$dV=-2.710147-3.506930$

$dV= -6.22$

Hence, the approximate change in volume is dV = -6.22 cubic units

8 0
3 years ago
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