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riadik2000 [5.3K]

3 years ago

12

One winter day, the temperature increased from a low -5 degrees F to a high of 40 degrees F.BY how many degrees did the temperat

ure change ?

1 answer:

sertanlavr [38]3 years ago

8 0

Answer:

The temperature changed by 45˚

Step-by-step explanation:

Set up an algebraic equation

-5 + x = 40

Add 5 to both sides

x = 45

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kupik [55]

Answer: 155y seconds

Step-by-step explanation:

Difference between 3y minutes and 25y seconds is written as

3y minutes- 25y seconds

First, Convert minutes to seconds = 60 seconds makes 1 minute, therefore, 3 minutes = 60 x 3= 180 seconds

Therefore 180y seconds - 25y seconds

= 155y seconds

I hope this helps, please mark as brainliest answer

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3 years ago

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Musya8 [376]

27 (rounded to the nearest year)

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3 years ago

I'm very confused <br> Please help ASAP!!!

harina [27]

Answer:

First case

0.3m = r

0.3m = 18

m = 18/0.3

m = 60 mins.

Second case

0.3m = r

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Third case

0.3m = r

0.3m = 72

m = 240 minutes

Fourth case

0.3m = r

0.3 x 360 = r

r = 108 rows

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4 0

4 years ago

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mel-nik [20]

Answer:

$\tan(\theta)=\frac{-\sqrt{5}}{2}$

Step-by-step explanation:

Since we are in quadrant 2, sine is positive. Since sine is positive and cosine is negative, then tangent is negative.

Now I'm going to find the sine value of this angle given using one of the Pythagorean Identities, namely $\sin^2(\theta)+\cos^2(\theta)=1$ .

If given $\cos(\theta)=\frac{-2}{3}$ , then we have $\sin^2(\theta)+(\frac{-2}{3})^2=1$ by substitution of $\cos(\theta)=\frac{-2}{3}$ .

Let's solve:

$\sin^2(\theta)+(\frac{-2}{3})^2=1$ for $\sin(\theta)$ .

$\sin^2(\theta)+\frac{4}{9}=1$

Subtract 4/9 on both sides:

$\sin^2(\theta)=1-\frac{4}{9}$

Simplify:

$\sin^2(\theta)=\frac{5}{9}$

Square root both sides:

$\sin(\theta)=\sqrt{\frac{5}{9}}$

$\sin(\theta)=\frac{\sqrt{5}}{\sqrt{9}}$

$\sin(\theta)=\frac{\sqrt{5}}{3}$

===========

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}=\frac{\frac{\sqrt{5}}{3}}{\frac{-2}{3}}$

Multiplying top and bottom by 3 gives:

$\tan(\theta)=\frac{\sqrt{5}}{-2}$

I'm going to move the factor of -1 to the top:

$\tan(\theta)=\frac{-\sqrt{5}}{2}$

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3 years ago

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sattari [20]

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3 years ago