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Anettt [7]
3 years ago
13

I don’t really understand this. can someone help

Mathematics
2 answers:
kari74 [83]3 years ago
8 0

Answer:

c

Step-by-step explanation:

melisa1 [442]3 years ago
6 0

Answer:

1

Step-by-step explanation:

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Trigonometry help?? Thank you so much!!
liq [111]

Answer:

7) a = 60; b = 100

8) c = 20

9) d = 45

Step-by-step explanation:

7)

a + 120 = 180

a = 60

b + 80 = 180

b = 100

8)

5c + 4c = 180

9c = 180

c = 20

9)

3d + d = 180

4d = 180

d = 45

6 0
3 years ago
The data given to the right includes data from 30 ​candies, and 5 of them are red. The company that makes the candy claims that
lina2011 [118]
100% expansión explanation
4 0
3 years ago
What is the expanded form of this number? 204.017
Anna [14]

Answer:

(2×100) + (4×1) + (1÷10) + (7÷100)

5 0
3 years ago
Please help me c:
netineya [11]

Oscar played games vs number of points he scored is, C) positive, linear association.

Step-by-step explanation:

  • no association is when points Oscar graph will remain between 8to10.
  • number of games he scored his points remain the same which is mean.
  • non linear is only when there is no straight line passing.
  • Linear is either exponential or polynomial.
  • Positive as the game increase he scoring abilities increases.
  • Negative as the game increases his scoring decreases.
  • Negative x axis will have more number of points.
  • Negative y axis will high to low of the graph.
  • Linear lines are best way to predict a data doesn't work will all data.

8 0
3 years ago
If L=√(x^2+y^2), dx/dt =-4, dy/dt=3, find dL/dt when x=4 and y=3
yulyashka [42]

Hello,

L=\sqrt{x^2+y^2} \\\\\dfrac{dx}{dt}=-4\\\\\dfrac{dy}{dt}=3\\\\x=4, y=3\\\\\dfrac{\partial L} {\partial x} =\dfrac{x}{\sqrt{x^2+y^2}} \\\\\dfrac{\partial L} {\partial y} =\dfrac{y}{\sqrt{x^2+y^2}} \\\\\dfrac{dL}{dt} =\dfrac{\partial L} {\partial x}*\dfrac{dx}{dt}+\dfrac{\partial L} {\partial y}*\dfrac{dy}{dt}\\\\=-4*\frac{-4}{\sqrt{4^2+3^2}} +3*\frac{3}{\sqrt{4^2+3^2}}\\\\=\dfrac{16}{5} +\dfrac{9}{5} \\\\=5\\

5 0
3 years ago
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