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sashaice [31]
3 years ago
15

Marshall left home biking at a constant rate of 20 mph. One hour later, it looked like it may rain, so his friend, brett, set ou

t in his car to pick him up. Brett drove at a constant rate of 60 m[h. How many hours will marshall have been riding his bike by the time brett catches up to him
Mathematics
1 answer:
Lana71 [14]3 years ago
4 0

Answer:

0.5

Step-by-step explanation:

First, we need to find how far ahead Marshall was.  Since he had been biking at 20 mph for one hour, he had gone 20 miles.

Next, we need to find how long it will take Brett to catch up to Marshall.  In order to do this, we need to find how much faster Brett is going than Marshall.  We do this by subtracting Marshall's speed from Brett's speed.

60 - 20 = 40.  So, Brett is catching up to Marshall at 40 mph.  Now, we figure out how long it will take for someone going 40 miles per hour to go 20 miles.  We find this by dividing 40 miles per hour by 20.  This is equal to 1/2 hour.  So, it will take Brett 0.5 hours to catch up to Marshall.  This is the same as A, so A is the correct answer.

We can check our answer by seeing how far Marshall and Brett will have gone.  Marshall will have been biking for 1.5 hours, so we multiply 20 * 1.5 = 30.  Marshall went 30 miles.

Brett drove for .5 hours at 60 mph, so he went 30 miles.  Since Brett and Marshall went the same distance, our answer is correct.

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Solve for X.<br> - 3x + 2 = -10x + 30
Nataly_w [17]

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Need this really quick plz
Oduvanchick [21]

Answer:

Let

x=sin-¹u

Sinx=u

let y=tan-¹v

tany=v

Substituting

Sin[x + y]

Applying the sine expansion

Sinxcosy + CosxSiny

Recall x =Sin-¹u

y=tan-¹v

Sin(Sin-¹u)Cos(tan-¹v) +Cos(sin-¹u)Sin(tan-¹v)

Now at this point

Here's what you do

For the first expression

Sin(Sin-¹u)

Let's simplify this

Let P = Sin-¹u

Taking sine of both sides

SinP=u

Draw a Right angled angle for this

Since Sine from SOHCAHTOA is OPP/HYP

Where P is the angle and u is the opposite and 1 is the hypotenuse since u is the same as u/1

substituting Sin-¹u = P

You have

Sin(Sin-¹u) = SinP

and from the triangle you drew

SinP = u

Taking the second express

Cos(tan-¹v)

Let Q=Tan-¹v

taking tan of both sides

tanQ=v

Draw a right angled triangle for this too

Since Tan from SOHCAHTOA is OPP/ADJ

Find the Hypotenuse cos you'll need it

Now Let's do the substitution again

We first said tan-¹v = Q

When we substitute it in Cos(tan-¹v)

We have CosQ

Cos Q from the second right angle triangle you drew is 1/√1+v²

Because CAH is adj/Hyp

So

the first part of the original Express

Which is

Sin(Sin-¹u)Cos(tan-¹v) is now simplified to

u(1/√1+v²).

Let's Move to the second part of the Original Expression

Cos(Sin-¹u)Sin(tan-¹v)

From our first solution

We said Sin-¹u= P

So replacing it here

we have Cos(sin-¹u) = CosP

let's leave the second one for now which is sin(tan-1v) We'll deal with this after the first

so Cos(Sin-¹u) = CosP

we can still use our first Right angle triangle for this because the angle was P.

so Cos P from that triangle will be

CosP= √1-u²

Now onto the next

Sin(tan-¹v)

From the Second solution of the first we did

we said let Tan-¹v =Q

Substituting this

we have

Sin(tan-¹v) = SinQ

using the second Right angle triangle because its angle is Q

We have

SinQ= v/√1+v²

Answer for second phase Which is

Cos(sin-¹u)Sin(tan-¹v) = √1-u²(v/√1+v²)

We're done

compiling our answers

The answer to

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You can still choose to factor out 1/√1+v² since it appears on both sides

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