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aliya0001 [1]
2 years ago
10

On a March day the temperature changed by an average of -0.5° C per hour from its high of 5° C at 1:00 pm. What was the temperat

ure at 10:00 pm?
Mathematics
1 answer:
pychu [463]2 years ago
5 0
0°? seems like it if you were to sub 0.5 ten times from 5
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The function y=-16t^2 +36 represents the height y (in feet) of a water droplet t seconds after falling from an icicle. After how
NikAS [45]

Answer:

1.5 seconds

Step-by-step explanation:

plug 0 in for y

0=-16t^2+36

then slove for t

-36=-16t^2

-36 / -16 = 9/4

9/4=t^2

square root of 9/4 is 3/2 or 1.5

t=1.5

3 0
2 years ago
Detetmine the next two terms in the sequence.
iragen [17]
Detetmine the next two terms in the sequence.
3, 16, 29, 42,
21, 25, 29, 33,
1, 2, 3, 5, 8, 1,


Determine the missing number in each sequence.

5,____, 10, 12 1/2
11, 5, 9, 4,____, 5, 2
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4 0
3 years ago
Where is the point at (2.5, -2.5) located
Vanyuwa [196]

Step-by-step explanation:

Go 2.5 units to the right from the origin (0,0) for your x coordinate

Go 2.5 units down from the origin (0,0) for your y coordinate

Where the lines meet is your point.

7 0
3 years ago
What is the slope of the line that passes through the points(7,-9) and (15,-29)?
Alex Ar [27]
(Y-Y1)/(X-X1)
(-9-(-29))/(7-15)
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3 0
3 years ago
Find sin (A-B) if sin A = 4/5 with A between 90 and 180 and if cos B = 3/5 with B between 0 and 90
quester [9]

Answer:

sin(A-B) = 24/25

Step-by-step explanation:

The trig identity for the differnce of angles tells you ...

sin(A -B) = sin(A)cos(B) -sin(B)cos(A)

We are given that sin(A) = 4/5 in quadrant II, so cos(A) = -√(1-(4/5)^2) = -3/5.

And we are given that cos(B) = 3/5 in quadrant I, so sin(B) = 4/5.

Then ...

sin(A-B) = (4/5)(3/5) -(4/5)(-3/5) = 12/25 + 12/25 = 24/25

The desired sine is 24/25.

3 0
3 years ago
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