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Anna [14]
4 years ago
8

At 3pm, the temperature outside was 5 1/5 degrees Fahrenheit. The temperature then fell steadily by 2 1/2 degrees per hour for t

he next four hours. What was the temperature at 7pm? Plz answer this I will give you 100 points
Mathematics
1 answer:
vazorg [7]4 years ago
8 0

Answer:

- 4 3/4 degrees Fahrenheit.

Step-by-step explanation

From 3 to 7 pm is 4 hours, so

Temperature at 7 pm =  5 1/4 - 4 * 2 1/2

=  21/4 - 4 * 5/2

= 21/4 - 10

=  21/4 - 40/4

= -19/4

= - 4 3/4 degrees F

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Therefore,

14x + 31x = 90 
45x = 90
x = 2

So we have a factor of two. Plug x back into the equation. 
The larger angle is 31 * 2 = 62 degrees.
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Pumpkin patch: Suppose we plant pumpkins in a garden where half of the plants are shaded. Then, to test a new fertilizer, we fer
VladimirAG [237]

Answer:

B. Fertilizer.

Step-by-step explanation:

In this experiment, you actually failed to include the fertilizer. This is because you are only fertilizing the pumpkins in the sun and not the shade. Therefore, you really do not know if the pumpkins that grew larger and prettier did so because of the fertilizer or the sun. What you should have done was either fertilized all pumpkins, none of the pumpkins, or half of the ones in the sun and half of the ones in the shade. That way you can correctly isolate the effect of the fertilizer on the pumpkins growth.

5 0
3 years ago
What do i have to look up to learn how to do this kind of trig. I went through kahn academy for trig but this stuff never came u
mariarad [96]

Answer:

  (e) √(x² -1)

Step-by-step explanation:

You can look up the Pythagorean theorem, and SOH CAH TOA.

__

<h3>Pythagorean theorem</h3>

This tells you the relations between the sides of a right triangle. For the purpose of finding tan(θ), you need to know the length of the side that is not marked on the diagram. It is found using the Pythagorean theorem.

The sum of squares of the legs is equal to the square of the hypotenuse. If we call the unknown leg "o", then ...

  1² +o² = x² . . . . . the Pythagorean relation

  o² = x² -1 . . . . . . subtract 1

  o = √(x² -1) . . . . take the square root

__

<h3>SOH CAH TOA</h3>

This mnemonic reminds you of the relationship between trig functions and side lengths of a right triangle.

  • Sin = Opposite/Hypotenuse
  • Cos = Adjacent/Hypotenuse
  • Tan = Opposite/Adjacent

This tells you that tan(θ) is the ratio of the side opposite θ to the side adjacent:

  tan(θ) = o/1 = o

From the previous section, o = √(x²-1), so ...

  tan(θ) = √(x² -1) . . . . . . . matches choice (e)

_____

<em>Additional comment</em>

Another way to think about this is in terms of the trig identities you have learned:

  sec(θ) = 1/cos(θ) = 1/(1/x) = x

  sec²(θ) = tan²(θ) +1   ⇒   tan(θ) = √(sec²(θ) -1) = √(x² -1)

7 0
2 years ago
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