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slava [35]
3 years ago
14

At 6:00 A.M. the temperature was 33°F. By noon the temperature had increased by 10°F and by 3:00 P.M. it had increased another 1

2°F. If at 10:00 P.M. the temperature had decreased by 15°F, how much does the temperature need to rise or fall to return to the original temperature of 33°F?
Mathematics
1 answer:
chubhunter [2.5K]3 years ago
3 0

So let's see here...


6:00 A.M. temperature is 33°F.

12:00 P.M temperature increased by 10°F making it 43°F.

3:00 P.M. temperature increased by another 12°F making it 55°F.

At 10:00 P.M it would decrease by 15°F making it 40°F.

The temperature would need to fall (or decrease) by 7°F to reach the original temperature of 33°F

Hope this helped!


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Are the area of a square and the length of its side directly proportional quantities?
Angelina_Jolie [31]

Answer:

Yes they are directly proportional quantities.

Step-by-step explanation:

We find the area of a square by;

A = length squared or (L)² , where 'L' stands for length and 'A' stands for area.

So Area = L²

Assume the length is a units and increase the length by 2 units

The initial area before increasing the length is a²

After increasing the length, the area becomes: (a + 2)² = a² + 4a + 4

Now we subtract the initial area from the final area and get;

(a² + 4a + 4) - a² = 4a + 4

So the new area increases by 4a + 4 units.

Hence, the area increases as the length increases implying that the area of a square is directly proportional to its length.

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A ∝ L

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A briefcase has a three-digit lock code that does not include zero as a digit.
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Solve each trigonometric equation such that 0 ≤ x ≤ 2????. Round answers to three decimal places.
zlopas [31]

Answer:

a) x= arccos(3/5) = 0.927

Just one solution

b) cos(x) = -\frac{5}{3}

But we know that the cosine can't be negative so then this equation no have solutions on the reals.

c) x = arcsin(\frac{1}{3})=0.3398

This is only the solution on the interval assumed.

d) x= -0.1145

Step-by-step explanation:

a. 5 cos(x) − 3 = 0

For this case we can do this:

5 cos(x) = 3

Now we can divide both sides by 5 and we got:

cos (x) = \frac{3}{5}

If we apply arccos on both sides we got:

x = arccos (\frac{3}{5})+2\pi n, x=2\pi-arccos (\frac{3}{5})+2\pi n

So for this case the possible solution is:

x= arccos(3/5) = 0.927

Just one solution. This is only the solution on the interval assumed.

b. 3 cos(x) + 5 = 0

We can do this:

3 cos (x) = -5

Then we can divide by 3 both sides and we got:

cos(x) = -\frac{5}{3}

But we know that the cosine can't be negative so then this equation no have solutions on the reals.

c. 3 sin(x) − 1 = 0

We can do this:

3 sin(x) = 1

Then we can divide both sides by 3 and we got:

sin(x) = \frac{1}{3}

The general solutions would be:

x = arcsin(\frac{1}{3}) + 2\pi n , x= \pi -arcsin (\frac{1}{3}) + 2\pi n

x = arcsin(\frac{1}{3})=0.3398

This is only the solution on the interval assumed.

d. tan(x) = −0.115

For this case we can solve the value of x like this:

x = arctan(-0.115) +\pi n

And then the only possible solution for this case is:

x= -0.1145

3 0
3 years ago
12.75 is 25% of what number
umka21 [38]
12.75 is 25% of 51
Hope this helps!!
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4 years ago
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