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ycow [4]
2 years ago
13

Write a system of linear equations for the problem below(you do not have to solve it) Twenty-six students attended the community

service event. The number of girls was 4 fewer than twice the number of boys. How many boys attended the community service event?
Mathematics
1 answer:
anzhelika [568]2 years ago
7 0

Answer:

Number of boys who attended the community service event = 10

Step-by-step explanation:

Let x denotes number of girls and y denotes number of boys.

Twenty-six students attended the community service event.

So,

x+y=26

The number of girls was 4 fewer than twice the number of boys.

x=2y-4

Put x=2y-4 in x+y=26

2y-4+y=26\\3y=26+4\\3y=30\\y=10

So,

Number of boys who attended the community service event = 10

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podryga [215]
The answer to your problem should be 5 wholes and 5/6 since the denominators are the same just add 6 more pieces to 4/6 to get an improper fraction then subtract to get the answer. Hope this helps and have a fabulous day!
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3 years ago
Find the mean of thw data 11,11,12,13,13,13,14,14,15,15,16,16,18
amid [387]

Answer:

the mean is 13

Step-by-step explanation:

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4 0
3 years ago
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

Sin(Sin-¹u - tan-¹v) = u(1/√1+v²) + [(√1-u²)(v/√1+v²)]

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

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Answer:

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