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Ksivusya [100]
3 years ago
7

A cell phone provider offers a plan that costs ​$40 per month plus ​$0.20 per text message sent or received. A comparable plan c

osts ​$60per month but offers unlimited text messaging. Complete parts a. and b. below. a. How many text messages would have to be sent or received in order for the plans to cost the same each​ month? In order for the plans to cost the​ same,____ text messages would have to be sent or receive
Mathematics
2 answers:
GalinKa [24]3 years ago
6 0
So what times .2 equals 20
100 messages
Ivahew [28]3 years ago
6 0
60 - 40 = 20
20 / .20 = 100
100 text messages would have to be sent or received. 
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Ahhhh help everyone help me solve this ....​
vovangra [49]

Answer:

y =  \frac{18 {w}^{2}  {x}^{2}  {z}^{2} }{25}

Step-by-step explanation:

\frac{3wx}{ \sqrt{2y} }  =  \frac{5}{2z}  \\  \\  \sqrt{2y}  \times 5 = 3wx \times 2z \\  \\  \sqrt{2y}   \times 5 = 6wxz \\  \\  \sqrt{2y}  =  \frac{6wxz}{5}  \\  \\  {( \sqrt{2y} })^{2}  =  {( \frac{6wxz}{5} })^{2}  \\  \\ 2y =  \frac{36 {w}^{2} {x}^{2} {z}^{2}   }{25}  \\  \\  \\  y =  \frac{36 {w}^{2} {x}^{2}  {z}^{2}  }{25}  \div 2 \\  \\  \\ y =  \frac{36 {w}^{2} {x}^{2} {z}^{2}   }{25}  \times  \frac{1}{2}  \\  \\  \\ y =  \frac{36 {w}^{2} {x}^{2}  {z}^{2}  }{50}  \\  \\  \\ y =  \frac{18 {w}^{2} {x}^{2}   {z}^{2} }{25}

I hope I helped you^_^

4 0
3 years ago
6x+32=−2x
velikii [3]
Whatever you do to one side, you have to do to the other.
Try to get all the x values and regular values together.
To do that, just subtract both sides by 6x to cancel it out on one side and subtract it from the other side:

6x-6x+32=-2x-6x
=
32=-8x
now divide both sides by -8 to get the x by itself

32/-8=-8/-8x
= 32/-8=x
-4=x
so your answer is: x= -4
5 0
3 years ago
A boat travels upstream for 48 miles in 3 hours and returns in 2 hours traveling downstream in a river. What is the rate of the
alekssr [168]

Answer:

20 mph and 4 mph

Step-by-step explanation:

GIVEN: A boat travels upstream for 48 miles in 3 hours and returns in 2 hours traveling downstream in a river.

TO FIND: What is the rate of the boat in still water and the rate of the current.

SOLUTION:

Let the speed of boat in still water be s and speed of current be a.

\therefore speed of water in upstream =s+a

  speed of water in downstream =s-a

As  speed=\frac{distance}{time}

s+a=\frac{48}{3}=16

s-a=\frac{48}{2}=24

Solving both we get

s=20\text{ mph}

a=4\text{ mph}

Hence speed of boat in still water and rate of current is 20mph and 4mph respectively.

3 0
3 years ago
I need help doing this can somebody please help?! It’s number 2.
stepladder [879]
Yes, if you solve both problems they come out to the same answer.
3 0
3 years ago
What is the answer to (6x1^-5)(8x10^-9)
baherus [9]
That is kinda confusing

6 0
3 years ago
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