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gregori [183]
2 years ago
9

Someone please help me

Mathematics
1 answer:
solmaris [256]2 years ago
6 0
37.
13 - 7 = 6
Her Mom Uses 6

38.
9 + 8 = 17
Andrew Had 17 Muffins
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To find the range you take the big number and subtract the small number.True False
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7 0
2 years ago
Math help please, thank you!
CaHeK987 [17]
The answer to the first question of the attached document is option 1. We obtain the answer subtracting the term n from the series with the term n-1.For example:
 -3 - (- 5) = 2
 -1 - (- 3) = 2
  1 - (- 1) = 2
 So you can see that the common difference is the 2.

 The answer to the second question is option 3:
 y = | x + 7 |
 We can confirm it by substituting values in the equation.
 For example:
 if we do y = 0 then x = -7
 if we do x = 0 then y = 7.
 As corresponds in the graph shown.
 Remember also that as a general rule yes to the equationy = | x | whose vertex is in the point (0,0) we add a positive real number "a" of form y = | x + a | then the graph of y = | x | will move "to" units in the negative direction of x.

 The answer to the third question is option 4.
 The quotient of x and "and" is constant.
 k = y / x
 Rewriting:
 y = kx
 You can see that it corresponds to the equation of a line that passes through the origin, this means that and is proportional to x and both vary directly
3 0
2 years ago
How do you do this and what’s the answer to this I been stuck on this question for litterly 20 mins
Vsevolod [243]

\bf T=2\pi \sqrt{\cfrac{L}{g}}\implies \stackrel{\textit{squaring both sides}}{(T)^2=\left( 2\pi \sqrt{\cfrac{L}{g}}\ \right)^2}\implies T^2=2^2\pi^2\sqrt{\left( \cfrac{L}{g} \right)^2} \\\\\\ T^2=4\pi^2\cfrac{L}{g}\implies T^2=\cfrac{4\pi^2L}{g}\implies gT^2=4\pi^2L\implies g=\cfrac{4\pi^2L}{T^2}

6 0
3 years ago
So confused. There is a link to the picture :)
user100 [1]
An = a1 + (n-1)d

an = 27 + (n-1)(-6)


4 0
3 years ago
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