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zloy xaker [14]
3 years ago
14

Find all solutions for a triangle with B=36 degrees, b=19, and c=36. Round to the nearest tenth.

Mathematics
1 answer:
lawyer [7]3 years ago
8 0

we know that

Applying the law of sines

b/sin B=c/sin C-------> solve for sin C

sin C=(c/b)*sin B-----> sin C=(36/19)*sin 36

sin C=1.1137-------> the value of the sines can not be greater than 1

The side lengths and angle given cannot be used to create a triangle

therefore


with the given values ​​there is no solution to form some triangle

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Aaron bought 8 red flags for the parade. Large flags cost $20 each. Medium flags cost $12 each. Aaron spent $112 in all. How man
andrew11 [14]

Answer:

Step-by-step explanation:

Price assuming he bought all large flags: 20*8=160

Amount of money more: 160-112=48

Difference of price in both flags: 20-12=8

Medium Flags: 48\div8=6

Large Flags: 8-6=2

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The reason why this works, is because the amount of money more comes from the difference between both flags.

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3 years ago
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ololo11 [35]

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3 years ago
Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used. Match each verbal description of a sequen
galben [10]

Answer:

I think the question is wrong so, I will try and explain with some right questions

Step-by-step explanation:

We are give 6 sequences to analyse

1. an = 3 · (4)n - 1

2. an = 4 · (2)n - 1

3. an = 2 · (3)n - 1

4. an = 4 + 2(n - 1)

5. an = 2 + 3(n - 1)

6. an = 3 + 4(n - 1)

1. This is the correct sequence

an=3•(4)^(n-1)

If this is an

Let know an+1, the next term

an+1=3•(4)^(n+1-1)

an+1=3•(4)^n

There fore

Common ratio an+1/an

r= 3•(4)^n/3•(4)^n-1

r= (4)^(n-n+1)

r=4^1

r= 4, then the common ratio is 4

Then

First term is when n=1

an=3•(4)^(n-1)

a1=3•(4)^(1-1)

a1=3•(4)^0=3.4^0

a1=3

The first term is 3 and the common ratio is 4, it is a G.P

2. This is the correct sequence

an=4•(2)^(n-1)

Therefore, let find an+1

an+1=4•(2)^(n+1-1)

an+1= 4•2ⁿ

Common ratio=an+1/an

r=4•2ⁿ/4•(2)^(n-1)

r=2^(n-n+1)

r=2¹=2

Then the common ratio is 2,

The first term is when n =1

an=4•(2)^(n-1)

a1=4•(2)^(1-1)

a1=4•(2)^0

a1=4

It is geometric progression with first term 4 and common ratio 2.

3. This is the correct sequence

an=2•(3)^(n-1)

Therefore, let find an+1

an+1=2•(3)^(n+1-1)

an+1= 2•3ⁿ

Common ratio=an+1/an

r=2•3ⁿ/2•(3)^(n-1)

r=3^(n-n+1)

r=3¹=3

Then the common ratio is 3,

The first term is when n =1

an=2•(3)^(n-1)

a1=2•(3)^(1-1)

a1=2•(3)^0

a1=2

It is geometric progression with first term 2 and common ratio 3.

4. I think this correct sequence so we will use it.

an = 4 + 2(n - 1)

Let find an+1

an+1= 4+2(n+1-1)

an+1= 4+2n

This is not GP

Let find common difference(d) which is an+1 - an

d=an+1-an

d=4+2n-(4+2(n-1))

d=4+2n-4-2(n-1)

d=4+2n-4-2n+2

d=2.

The common difference is 2

Now, the first term is when n=1

an=4+2(n-1)

a1=4+2(1-1)

a1=4+2(0)

a1=4

This is an arithmetic progression of common difference 2 and first term 4.

5. I think this correct sequence so we will use it.

an = 2 + 3(n - 1)

Let find an+1

an+1= 2+3(n+1-1)

an+1= 2+3n

This is not GP

Let find common difference(d) which is an+1 - an

d=an+1-an

d=2+3n-(2+3(n-1))

d=2+3n-2-3(n-1)

d=2+3n-2-3n+3

d=3.

The common difference is 3

Now, the first term is when n=1

an=2+3(n-1)

a1=2+3(1-1)

a1=2+3(0)

a1=2

This is an arithmetic progression of common difference 3 and first term 2.

6. I think this correct sequence so we will use it.

an = 3 + 4(n - 1)

Let find an+1

an+1= 3+4(n+1-1)

an+1= 3+4n

This is not GP

Let find common difference(d) which is an+1 - an

d=an+1-an

d=3+4n-(3+4(n-1))

d=3+4n-3-4(n-1)

d=3+4n-3-4n+4

d=4.

The common difference is 4

Now, the first term is when n=1

an=3+4(n-1)

a1=3+4(1-1)

a1=3+4(0)

a1=3

This is an arithmetic progression of common difference 4 and first term 3.

5 0
3 years ago
Plz help LOL <br> 5(x-3)-x=
V125BC [204]
5(x-3)-x =
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3 years ago
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ZanzabumX [31]

Answer:

Step-by-step explanation:

Just truely be open about your feelings, don't try and sugarcoat it, or leave things out, just be honest with her and tell her straight up, politely of course.

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2 years ago
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