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3 years ago

9

How do you write the nth term rule for the sequence 1 , 3 , 5 , 7

1 answer:

Rashid [163]3 years ago

5 0

basically the answer is 2n-1 2n comes from the difference to get from 1 to three and the -1 from 2n to get to 1 you need to minus 1 so you get -1

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What is the reminder when (3x^3-2x^2+4x-3) is divided by (x^2+3x+3)

NARA [144]

(For full answer you might have to go to the comments)

Answer: 28x+30

Explanation: we divide (3x^3-2x^2+4x-3) by (x^2+3x+3) Using long division
3x-11
___________________
(x^2+3x+3) 3x^3-2x^2+4x-3

-(3x^3+9x^2+9x)
__________________

-11x^2-5x-3

-(-11x^2-33x-33)
____________
28x+30

So our remainder will be 28x+30

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3 years ago

Evaluate x + 8 when x = 7<br><br> a. 15<br><br> b. 1<br><br> c. -1<br><br> d. 56

Tems11 [23]

Answer:

a) 15

Step-by-step explanation:

x+8

substitute x value in

7+8

=15

8 0

3 years ago

Read 2 more answers

Find the parametric equations for the line that is tangent to the given curve at the given parameter (3cost)i (t^4 -3sint)j

ale4655 [162]

Answer:

Following are the solution to this question:

Step-by-step explanation:

Please find the complete question in the attached file.

In the given equation, when the point t=0

So,

$\to r(0) = (3 \cos 0)i + (0^4 - 6 \sin 0)j + (2e^{3\times 0})k)$

$= (3 \times 1)i + (0 - 0)j + (2e^{0})k)\\\\ = 3i + 0j + (2 \times 1)k)\\\\ = 3i + 0j + 2k \\$

The value of the coordinates are $3, 0, 2$ . so, the equation of the line is:

$\to \frac{(x-3)}{3 \cos \ t} = \frac{(y-0)}{(t^{4}-6 \sin \ t)} = \frac{(z-2)}{2e^{3t}}=k$

5 0

3 years ago

9. Alicia Martin's savings account has a principle of $1,200. It earns 6% interest compounded quartly

zavuch27 [327]

Answer:

$9) \$1236.27\,10)\,\$6451.07\, 11)\,\$10,152.87 \,12)\,\$907.95 \,13)\,\$4957.69$

Step-by-step explanation:

9) Since Alicia Martin's savings earns 6% quarterly for two quarters then:

$A=P(1+\frac{r}{n})^{nt}$ ⇒ Amount (A), Principle (P), rate (r) in decimal form, number of compoundings (n) a year and t, in year or its fractions.

$A=P(1+\frac{r}{n})^{nt}\Rightarrow A=1200(1+\frac{0.06}{4})^{4*\frac{1}{2}}\Rightarrow A=\$1236.27$

10) Aubrey Daniel's case:

$A=P(1+\frac{r}{n})^{nt}\Rightarrow A=5725(1+\frac{0.04}{4})^{4*3}\Rightarrow A\approx \$6451.07$

11) As for Angelo, similarly to Alicia.

$A=P(1+\frac{r}{n})^{nt}\Rightarrow A=9855(1+\frac{0.06}{4})^{4*\frac{1}{2}}\Rightarrow A\approx \$10,152.87$

12) Simpson's. For semiannual n=2

$A=P(1+\frac{r}{n})^{nt}\Rightarrow A=860(1+\frac{0.055}{2})^{2*1}\Rightarrow A\approx \$907.95$

13) Jana Lacey amount:

$A=P(1+\frac{r}{n})^{nt}\Rightarrow A=4860(1+\frac{0.04}{4})^{4*\frac{1}{2}}\Rightarrow A\approx \$4957.69$

6 0

3 years ago

Use long division to find the quotient below (8x^3+20x^2+36)/(2x+6)

blsea [12.9K]

Answer: 14x - 42

Hope this helps!

6 0

1 year ago