5) Find the missing term in this proportion. 30 360 - 12

An airplane flies with a constant speed of 620 miles per hour. How long will it take to

LekaFEV [45]

Answer:

30 minutes

Step-by-step explanation:

Alright lets start by laying out what the problem gives us :

A plane flies 620 mph.

It is asking how long it will take to travel 310 miles.

We need to give our answer in minutes or hours.

So lets take mph and look at the full meaning "Miles per hour"

An hour has 60 minutes in it.

So really it is saying miles per 60 minutes as well as miles per hour

So you could say the plane is traveling 620 miles every 60 minutes.

So now lets look at the number 310.

Just looking at it I can tell you that 310 looks very close to half of 620, in fact it is exactly half!

So lets divide the 620 mph by the 310 miles it is asking and we get 2! Proving that 310 is half of 620.

Alright now lets divide both the number of miles and number of minutes :

620 miles per 60 minutes

divided by 2 both miles and minutes

310 miles per 30 minutes

So this is your answer. After this you get that it would take 30 minutes to travel 310 miles in a plane going 620 miles per hour!

I hope this helps and I could help you understand it. :)

If you have any questions feel free to comment or dm me.

7 0

2 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

Factorise 3(m-4)+b(4-m)

Maurinko [17]

Answer:

-(m-4)(b-3)

Step-by-step explanation:

Add and subtract the second term to the expression and factor by grouping

8 0

2 years ago

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Please help with number 3!!! ASAP!!!!

Otrada [13]

LCM I believe. Because it’s every 6 days he does both

3 0

3 years ago

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X^2-2x+24 solve for x (factor)

Naddika [18.5K]

The correct answer is either 6 or -4

8 0

2 years ago