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Alex787 [66]
3 years ago
7

On Halloween you collected 132 pieces of candy.Every day you brought 2 pieces of candy to your best friend and you have 30 left.

How many days have you been sharing your candy
Mathematics
2 answers:
Alex777 [14]3 years ago
8 0

Answer:

51 days

Step-by-step explanation:

You would subtract 30 from 132 to get 102 pieces of candy that you have shared. If you share two a day, then divide 102 pieces by 2 to get 51 days :)

You can check by doing 2*51=102 + 30 leftover= 132 pieces

Verizon [17]3 years ago
6 0

Answer:

100

Step-by-step explanation:

Becuase that is 2 everyday so it should be between 100 and 50 if not I'm sorry if it's wrong but it should be right

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PLEASE HELP!!<br><br><br><br><br> Ill take what ever i can get!
Marina86 [1]

the evidence does not support the argument because it addresses the bookmobile than the library itself

7 0
3 years ago
Read 2 more answers
Abigail flipped a coin 20 times, and it landed with heads facing up 13 times. What percent of the coin flips landed with heads f
DIA [1.3K]
65% of the coins flipped landed with heads facing up because 13 were faced up out of 20.

Divide:
13/20=0.65=65%

Hope this helped☺☺
5 0
3 years ago
In New York City at the spring equinox there are 12 hours 8 minutes of daylight. The
slamgirl [31]

Answer:

  independent: day number; dependent: hours of daylight

  d(t) = 12.133 +2.883sin(2π(t-80)/365.25)

  1.79 fewer hours on Feb 10

Step-by-step explanation:

a) The independent variable is the day number of the year (t), and the dependent variable is daylight hours (d).

__

b) The average value of the sinusoidal function for daylight hours is given as 12 hours, 8 minutes, about 12.133 hours. The amplitude of the function is given as 2 hours 53 minutes, about 2.883 hours. Without too much error, we can assume the year length is 365.25 days, so that is the period of the function,

March 21 is day 80 of the year, so that will be the horizontal offset of the function. Putting these values into the form ...

  d(t) = (average value) +(amplitude)sin(2π/(period)·(t -offset days))

  d(t) = 12.133 +2.883sin(2π(t-80)/365.25)

__

c) d(41) = 10.34, so February 10 will have ...

  12.13 -10.34 = 1.79

hours less daylight.

5 0
3 years ago
Please help someone
DiKsa [7]

The first equation is x = -1

The second and third equations are no solution

The third equation is all rel numbers.


We can tell each one by solving them. In the first one, you get the following.


4 + x = -8x - 5 ----> Add 8x to both sides

4 + 9x = -5 ----> Subtract 4 from both sides

9x = -9 ----> Divide both sides by 9

x = -1


For the middle two, when you attempt to solve, you get untrue statements. This shows there are no solutions. See the example below.


7 + 2x = 2x - 7 ----> Subtract 2x from both sides

7 = -7 (UNTRUE)


And for the last one, each term cancels out, which shows that we have all real solutions.


-3x + 3 = 3( 1 - x) ----> Distribute the 3

-3x + 3 = 3 - 3x ----> Add 3x to both sides

3 = 3 (TRUE)

8 0
3 years ago
For the following parameterized​ curve, find the unit tangent vector T​(t) at the given value of t. r​(t) = &lt; 8 t,10,3 sine 2
san4es73 [151]

Answer:

The tangent vector for t = 0 is:

\vec T (t) = \left \langle \frac{8}{10}, 0, \frac{6}{10}   \right\rangle

Step-by-step explanation:

The function to be used is \vec r(t) = \langle 8\cdot t, 10, 3\cdot \sin (2\cdot t)\rangle

The unit tangent vector is the gradient of \vec r (t) divided by its norm, that is:

\vec T (t) =  \frac{\vec \nabla r (t)}{\|\vec \nabla r (t)\|}

Where \vec \nabla is the gradient operator, whose definition is:

\vec \nabla f (x_{1}, x_{2},...,x_{n}) = \left\langle \frac{\partial f}{\partial x_{1}}, \frac{\partial f}{\partial x_{2}},...,\frac{\partial f}{\partial x_{n}} \right\rangle

The components of the gradient function of \vec r(t) are, respectively:

\frac{\partial r}{\partial x_{1}} = 8, \frac{\partial r}{\partial x_{2}} = 0 and \frac{\partial r}{\partial x_{3}} = 6 \cdot \cos (2\cdot t)

For t = 0:

\frac{\partial r}{\partial x_{1}} = 8, \frac{\partial r}{\partial x_{2}} = 0 and \frac{\partial r}{\partial x_{3}} = 6

The norm of the gradient function of \vec r (t) is:

\| \vec \nabla r(t) \| = \sqrt{8^{2}+0^{2}+ [6\cdot \cos (2\cdot t)]^{2}}

\| \vec \nabla r(t) \| = \sqrt{64 + 36\cdot \cos^{2} (2\cdot t)}

For t = 0:

\| \vec r(t) \| = 10

The tangent vector for t = 0 is:

\vec T (t) = \left \langle \frac{8}{10}, 0, \frac{6}{10}   \right\rangle

8 0
3 years ago
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