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lorasvet [3.4K]
2 years ago
9

Convert the angle 0 = 70° to radians.​

Mathematics
1 answer:
xxMikexx [17]2 years ago
3 0

Answer:

7/18π radians   Exact answer

1.22 radians      Decimal approximation

Step-by-step explanation:

π radians  = 180º

70 * π/180 = 7/18π

7/18 * 3.14159 ≈ 1.22

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Find the amplitude y = -5 sin one half x
Alex_Xolod [135]

Answer:

amplitude=5

Step-by-step explanation:

y=a sin x

amplitude =|a|

here a=-5

|a|=5

8 0
3 years ago
If your data set does not include an outlier which measure of center, Mean or Median, is best to use to describe your data. Expl
VLD [36.1K]
When there isn't an outlier, the mean is the best measure of central tendency, because it gives you a more arithmetic answer of the true center since all numbers are included in the equation rather than at max 2 numbers in the middle of the list.
3 0
3 years ago
K= 35.50x + 42<br> k=991x + 27
SSSSS [86.1K]

Answer:

k=42.5573 and x=0.015699

Step-by-step explanation:

Step: Solve k=35.5x+42for k:

k=35.5x+42

Step: Substitute 35.5x+42 for k in k=991x+27:

k=991x+27

35.5x+42=991x+27

35.5x+42+−991x=991x+27+−991x(Add -991x to both sides)

−955.5x+42=27

−955.5x+42+−42=27+−42(Add -42 to both sides)

−955.5x=−15

−955.5x

−955.5

=

−15

−955.5

(Divide both sides by -955.5)

x=0.015699

Step: Substitute0.015699 for x in k=35.5x+42:

k=35.5x+42

k=(35.5)(0.015699)+42

k=42.5573(Simplify both sides of the equation)

Hope this helps!

:)

6 0
2 years ago
Solve for n.<br>--42 – 6n = -30​
rodikova [14]
N=-2
Add 42 to both sides to get -6n=12
Divide both sides by -6
7 0
3 years ago
hich function has a range of {y|y ≤ 5}? f(x) = (x – 4)2 + 5 f(x) = –(x – 4)2 + 5 f(x) = (x – 5)2 + 4 f(x) = –(x – 5)2 + 4
Brums [2.3K]

Answer:

f(x)=-(x-4)^2+5

Step-by-step explanation:

The function f(x)=-(x-4)^2+5 has  the vertex at (4,5).


Since a=-1\:, the graph of the function is a maximum graph.


The highest y-value is the y-coordinate of the vertex which is 5.


Therefore the range is {y|y\le 5}


Therefore the correct answer is B.


6 0
3 years ago
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