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Sedaia [141]
3 years ago
13

The path of his motorcycle was given approximately by H=-0.005x^2+2.39x+600 where H was measured in ft above the river and x was

the distance from his launch ramp. How high above the river was the launch ramp? I found that the maximum height was 600ft and the horizontal distance was 234ft.​
Mathematics
1 answer:
Simora [160]3 years ago
3 0

Answer:

H = 600 ft.

H_{max} =885.6 ft at x =239 ft.

Step-by-step explanation:

The path of a motorcycle is given by H =-0.005x^{2}+2.39x + 600 .....(1) where H is the height above the river in ft and x is the distance from his launch camp.

Putting x = 0, the height of the launch camp from the river is H = 600 ft.

Now. differentiating equation (1), with respect to x both sides we get,

\frac{dH}{dx} = 0.01x -2.39 =0 {Condition for H to be maximum is \frac{dH}{dx} =0}

⇒ x = 239 ft.

So, H_{max} = -0.005 (239)^{2} + 2.39 \times 239 + 600 =885.6 ft. (Answer)

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Nadia is a stockbroker. She earns 12% commission each week. Last week, she sold $5400 worth of stocks. How much did she make las
Inga [223]

Answer:

a). Nadia made a total of $825 last week in commission sales

b). The total amount Nadia in commission in 2011=$42,900

Step-by-step explanation:

Step 1

Express the amount in commission she earns from her sales as follows;

C=S×r

where;

C=commission earned

S=total value of stock sales

r=commission rate

In our case;

C=unknown

S=$7,500

r=11%=11/100=0.11

replacing;

C=11% of 7,500

C=(11/100)×7,500

C=0.11×7,500=825

Nadia made a total of $825 last week in commission sales

b). To express the total amount she made in 2011, we derive the formula below;

Total amount (2011)=Average weekly commission×number of weeks in 2011

where;

Average weekly commission=$825

number of weeks in 2011=52

replacing;

Total amount (2011)=(825×52)=42,900

The total amount Nadia in commission in 2011=$42,900

6 0
3 years ago
2m - 6 = 8m then 3m =
Vinil7 [7]

Answer:

3m = -3

Step-by-step explanation:

You are given

2m - 6 = 8m,

hence,

2m - 8m = 6,

-6m = 6,

m = 6%2F%28-6%29,

m = -1.

Therefore, 3m = -3.

5 0
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
12 grapes cost 90 $how many can bought in 40$
likoan [24]

Let x = the amount of grapes that can be bought for 0.40.


Note: By 90 I assume you mean 90 cents. The same for 40.


12/x = 0.90/0.40


0.90x = 12(0.40)


0.90x = 4.80


x = 4.80 ÷ 0.90


x = 5.33333


We can round the decimal to the ones place to get 5 grapes.




4 0
3 years ago
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25 dollars for each hour. 
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