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4 years ago

Will mark brainliest if right and show your work...

Mathematics

1 answer:

skelet666 [1.2K]4 years ago

5 0

(9x^4-13x^3-x-7)+(7x^3-2x+1)

Sort the terms by their powers, as x's with the same powers can be added together:

9x^4

-13x^3 + 7x^3 = -6x^3

-x - 2x = -3x

-7 +1 = -6

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The table below shows the value of Henry's car at the end for each of the first 3 years after it is purchased. The values form a

SVETLANKA909090 [29]

Using a geometric sequence, it is found that the approximate value of the car at the end of the 10th year will be given by:

A. $6,974.

<h3>What is a geometric sequence?</h3>

A geometric sequence is a sequence in which the result of the division of consecutive terms is always the same, called common ratio q.

The nth term of a geometric sequence is given by:

$a_n = a_1q^{n-1}$

In which $a_1$ is the first term.

In this problem, the first term and the common ratio are given, respectively, by:

$a_1 = 18000, q = \frac{16200}{18000} = 0.9$

Hence the equation is:

$a_n = 18000(0.9)^{n-1}$

At the end of the 10th year, the value will be of:

$a_{10} = 18000(0.9)^{10-1} = 6974$

Hence option A is correct.

More can be learned about geometric sequences at brainly.com/question/11847927

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6 0

2 years ago

HURRYYYY

kati45 [8]

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3 0

4 years ago

Read 2 more answers

Each week, Kathy receives a batch of phones that need repairs. The number of phones that she has left to fix at the end of each

adell [148]

Answer:

A.) Kathy starts each week with 108 phones to fix.

6 0

4 years ago

The price of a bicycle was reduced from $120 to $90. By what percentage was the price of the bicycle reduced?

Degger [83]

Answer:price of the bicycle was reduced by 25%

Step-by-step explanation:

Step 1

Old Price of bicycle = $120

New Price of bicycle= $90

Price decrease= $120-$90= $30

Step 2

Percentage decrease = Decrease / Original price x 100

= (30/120) x 100

1/4 x 100

= 25%

Therefore the percentage decrease in the price is 25%

5 0

3 years ago

Consider the following vector function. R(t) = 9 2 t, e9t, e−9t (a) find the unit tangent and unit normal vectors t(t) and n(t)

garik1379 [7]

The unit tangent vector is T(u) and the unit normal vector is N(t) if the vector function. R(t) is equal to 9 2 t, e9t, e−9t.

<h3>What is vector?</h3>

It is defined as the quantity that has magnitude as well as direction also the vector always follows the sum triangle law.

We have vectored function:

$\rm R(t) = (9\sqrt{2t}, e^{9t}, e^{-9t})$

Find its derivative:

$\rm R'(t) = (9\sqrt{2}, 9e^{9t}, -9e^{-9t})$

Now its magnitude:

$\rm |R'(t) |= \sqrt{(9\sqrt{2})^2+ (9e^{9t})^2+ (-9e^{-9t})^2}$

After simplifying:

$\rm R'(t) = 9 \dfrac{e^{18t}+1}{e^{9t}}$

Now the unit tangent is:

$\rm T(u) = \dfrac{R'(t)}{|R'(t)|}$

After dividing and simplifying, we get:

$\rm T(u) = \dfrac{1}{e^{18t}+1} (\sqrt{2}e^{9t}, e^{18t}, -1)$

Now, finding the derivative of T(u), we get:

$\rm T'(u) = \dfrac{1}{(e^{18t}+1)^2} (9\sqrt{2}e^{9t}(1-e^{18t}), 18e^{18t}, 18e^{18t})$

Now finding its magnitude:

$\rm |T'(u) |= \dfrac{1}{(e^{18t}+1)^2} (9\sqrt{2}e^{9t}(1-e^{18t})^2+ (18e^{18t})^2+( 18e^{18t})^2)$

After simplifying, we get:

$\rm |T'(u)|= \dfrac{9\sqrt{2}e^{9t}}{e^{18t}+1}$

Now for the normal vector:

Divide T'(u) and |T'(u)|

We get:

$\rm N(t) = \dfrac{1}{e^{18t}+1} ( 1-e^{18t}, \sqrt{2}e^{9t}, \sqrt{2}e^{9t})$

Thus, the unit tangent vector is T(u) and the unit normal vector is N(t) if the vector function. R(t) is equal to 9 2 t, e9t, e−9t.

Learn more about the vector here:

brainly.com/question/8607618

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3 0

2 years ago