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2 years ago

What value, written as a decimal, should Lena use as the common ratio?

Mathematics

1 answer:

inn [45]2 years ago

7 0

Answer:

an=1*2.5^(n-1)

=2.5^(n-1)

Step-by-step explanation:

Complete question below:

What value, written as a decimal, should Lena use as the common ratio? Lena is asked to write an explicit formula for the graphed geometric sequence. On a coordinate plane, 3 points are plotted. The points are (1, 1), (2, 2.5), (3, 6.25).

Solution

Point (1, 1), (2, 2.5), (3, 6.25).

a=1

ar=2.5

ar^2=6.25

From ar and ar^2

r=6.25/2.5

=2.5

r=2.5

an=ar^(n-1)

Therefore, the explicit formula is

an=1*2.5^(n-1)

=2.5^(n-1)

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3. Tickets to a show cost $5.50 for adults and 54.25 for students. A family is purchasing2

bulgar [2K]

Step-by-step explanation:The cost of an adult ticket to the show is $5.5

The cost of a student ticket to the show is $4.25

A family is purchasing 2 adult tickets and 3 student tickets. This means that the total cost of 2 adult tickets would be

2 × 5.5 = $11

It also means that the total cost of 3 student tickets would be

3 × 4.25 = $12.75

The total cost of 2 adult tickets and 3 student tickets would be

11 + 12.75 = $23.75

If the family pays $25, the exact amount of change they should receive is

25 - 23.75 = $1.25

7 0

2 years ago

Cos (90-theta) • cosec90-theta) =tano. How?

denis-greek [22]

Answer:

______________________________________________________

TRIGONOMETRY IDENTITIES TO BE USED IN THE QUESTION :-

For any right angled triangle with one angle α ,

$\cos (90 - \alpha ) = \sin \alpha$ or $\sin(90 - \alpha ) = \cos\alpha$

$cosec \: (90 - \alpha ) = \sec\alpha$ or $\sec(90 - \alpha ) = cosec\:\alpha$

SOME GENERAL TRIGNOMETRIC FORMULAS :-

$\sin \alpha = \frac{1}{cosec \: \alpha }$ or $cosec \: \alpha = \frac{1}{\sin \alpha }$

$\cos \alpha = \frac{1}{\sec \alpha }$ or $\sec \alpha = \frac{1}{\cos \alpha }$

______________________________________________________

Now , lets come to the question.

In a right angled triangle , let one angle be α (in place of theta) .

So , lets solve L.H.S.

$\cos (90 - \alpha ) \times cosec(90 - \alpha )$

$=> sin\alpha \times \sec\alpha$

$=> \sin\alpha \times \frac{1}{\cos\alpha }$

$=> \frac{\sin\alpha }{\cos\alpha }$

$=> \tan\alpha$ = R.H.S.

∴ L.H.S. = R.H.S. (Proved)

3 0

2 years ago

Where should the point P be chosen on line segment AB so as to maximize the angle θ? (Assume a = 4 units, b = 5 units, and c = 9

taurus [48]

From the figure, let the distance of point P from point A on line segment AB be x and let the angle opposite side a be M and the angle opposite side c be N.

Using pythagoras theorem,
$\tan M= \frac{a}{b-x} \\ \\ M=\tan^{-1}\left(\frac{a}{b-x}\right)$
and
$\tan N= \frac{c}{x} \\ \\ N=\tan^{-1}\left(\frac{c}{x}\right)$

Angle θ is given by
$\theta=180-M-N \\ \\ =180-\tan^{-1}\left(\frac{a}{b-x}\right)-\tan^{-1}\left(\frac{c}{x}\right)$

Given that a = 4 units, b = 5 units, and c = 9 units, thus
$\theta=180-\tan^{-1}\left(\frac{4}{5-x}\right)-\tan^{-1}\left(\frac{9}{x}\right)$

To maximixe angle θ, the differentiation of θ with respect to x must be equal to zero.
i.e.
$\frac{d\theta}{dx} = -\frac{4}{x^2-10x+41} + \frac{9}{x^2+81} =0 \\ \\ -4(x^2+81)+9(x^2-10x+41)=0 \\ \\ -4x^2-324+9x^2-90x+369=0 \\ \\ 5x^2-90x+45=0 \\ \\ x^2-18x+9=0 \\ \\ x=9\pm6 \sqrt{2}$

Given that x is a point on line segment AB, this means that x is a positive number less than 5.

Thus
$x=9-6 \sqrt{2}=0.5147$

Therefore, The distance from A of point P, so that angle θ is maximum is 0.51 to two decimal places.

6 0

3 years ago

4x + 5y = 10 Question: A. Convert the equation to slope intercept form and justify each step. B. Explain how to graph the linear

VladimirAG [237]

The answer to your question is A but i dont see a D option so it might be wrong

8 0

3 years ago

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Given the function h(x)= 5x-8, find the value of r such that h(3-r) = 6r

Keith_Richards [23]

the answer is r= 3h/h+6

4 0

2 years ago