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3 years ago

Answer the following questions using the table below.

Mathematics

1 answer:

irina1246 [14]3 years ago

3 0

Answer:

the - are the x or y

Step-by-step explanation:

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A weather station on the top of a mountain reports that the temperature is currently 0 ∘ C and has been falling at a constant ra

LuckyWell [14K]

Answer:

After 2 hours: -6°C.

After 5 hours: -15°C.

After $\frac{1}{2}$ hours: -1.5°C

1 hour ago: 3°C.

3 hours ago: 9°C.

4.5 hours ago: 13.5°C.

To find these values, you simply need to multiply the hours elapsed by 3°C. Subtract from 0°C for future temperatures, and ADD to 0°C for the past.

Let's say you are solving for the temperature in 5 hours. All you need to do is:

3°C× 5= 15°C. 0°C-15°C= -15°C

For the temperature 3 hours AGO, what you will need to do instead is:

3°C×3= 9°C. 0°C+ 9°C= 9°C.

8 0

3 years ago

Tell whether the difference between two negative integers is always, sometimes, or never positive. The difference between two ne

babymother [125]

Answer:

. 2) It’s positive only if the first integer is greater

3 0

3 years ago

Read 2 more answers

If c is between a and b then Ac= CB

rodikova [14]

It only works if C is the midpoint of AB

5 0

4 years ago

Read 2 more answers

Write the number 33×10^-3 in scientific notation

cluponka [151]

Answer:

Step-by-step explanation:

33×10^-3 is equivalent to 33/1000, or 0.033.

This, in turn, is equivalent to 3.3×10^(-2) (which is in scientific notation)

4 0

3 years ago

Derive these identities using the addition or subtraction formulas for sine or cosine: sinacosb=(sin(a+b)+sin(a-b))/2

Sergeu [11.5K]

Answer:

The work is in the explanation.

Step-by-step explanation:

The sine addition identity is:

$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$ .

The sine difference identity is:

$\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(a)$ .

The cosine addition identity is:

$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$ .

The cosine difference identity is:

$\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$ .

We need to find a way to put some or all of these together to get:

$\sin(a)\cos(b)=\frac{\sin(a+b)+\sin(a-b)}{2}$ .

So I do notice on the right hand side the $\sin(a+b)$ and the $\sin(a-b)$ .

Let's start there then.

There is a plus sign in between them so let's add those together:

$\sin(a+b)+\sin(a-b)$

$=[\sin(a+b)]+[\sin(a-b)]$

$=[\sin(a)\cos(b)+\cos(a)\sin(b)]+[\sin(a)\cos(b)-\cos(a)\sin(b)]$

There are two pairs of like terms. I will gather them together so you can see it more clearly:

$=[\sin(a)\cos(b)+\sin(a)\cos(b)]+[\cos(a)\sin(b)-\cos(a)\sin(b)]$

$=2\sin(a)\cos(b)+0$

$=2\sin(a)\cos(b)$

So this implies:

$\sin(a+b)+\sin(a-b)=2\sin(a)\cos(b)$

Divide both sides by 2:

$\frac{\sin(a+b)+\sin(a-b)}{2}=\sin(a)\cos(b)$

By the symmetric property we can write:

$\sin(a)\cos(b)=\frac{\sin(a+b)+\sin(a-b)}{2}$

3 0

3 years ago