All categories

3 years ago

Select the recursive rule that models the sequence below.14,20,26,32

Mathematics

1 answer:

AURORKA [14]3 years ago

5 0

Answer:

a₁ = 14, aₙ = 14 + 6(n - 1)

Step-by-step explanation:

<u>Sequence given:</u>

14, 20, 26, 32

<u>We see common difference is 6:</u>

32-26 = 26-20 = 20 - 14 = 6

<u>Recursive formula is:</u>

a₁ = 14,
aₙ = 14 + 6(n - 1)

You might be interested in

What kind of angles are 1 and 5 in the picture below?

9966 [12]

Right angle because you don’t have the full thing

4 0

3 years ago

If the area of the region bounded by the curve y^2 =4ax and the line x= 4a is 256/3 Sq units, using integration find the value o

almond37 [142]

If the area of the region bounded by the curve $y^2 =4ax$ and the line $x= 4a$ is $\frac{256}{3}$ Sq units, then the value of $a$ will be $2$ .

<h3>What is area of the region bounded by the curve ?</h3>

An area bounded by two curves is the area under the smaller curve subtracted from the area under the larger curve. This will get you the difference, or the area between the two curves.

Area bounded by the curve $=\int\limits^a_b {x} \, dx$

We have,

$y^2 =4ax$

⇒ $y=\sqrt{4ax}$

$x= 4a$ ,

Area of the region $=\frac{256}{3}$ Sq units

Now comparing both given equation to get the intersection between points;

$y^2=16a^2$

$y=4a$

So,

Area bounded by the curve $= \[ \int_{0}^{4a} y \,dx \]$

$\frac{256}{3} =\[ \int_{0}^{4a} \sqrt{4ax} \,dx \]$

$\frac{256}{3}= \[\sqrt{4a} \int_{0}^{4a} \sqrt{x} \,dx \]$

$\frac{256}{3}= \[2\sqrt{a} \int_{0}^{4a} x^{\frac{1}{2} } \,dx \]$

$\frac{256}{3}= 2\sqrt{a} \left[\begin{array}{ccc}\frac{(x)^{\frac{1}{2}+1 } }{\frac{1}{2}+1 }\end{array}\right] _{0}^{4a}$

$\frac{256}{3}= 2\sqrt{a} \left[\begin{array}{ccc}\frac{(x)^{\frac{3}{2} } }{\frac{3}{2} }\end{array}\right] _{0}^{4a}$

$\frac{256}{3}= 2\sqrt{a} *\frac{2}{3} \left[\begin{array}{ccc}(x)^{\frac{3}{2}\end{array}\right] _{0}^{4a}$

On applying the limits we get;

$\frac{256}{3}= \frac{4}{3} \sqrt{a} \left[\begin{array}{ccc}(4a)^{\frac{3}{2} \end{array}\right]$

$\frac{256}{3}= \frac{4}{3} \sqrt{a} *\sqrt{(4a)^{3} }$

$\frac{256}{3}= \frac{4}{3} \sqrt{a} * 8 *a^{2} \sqrt{a}$

$\frac{256}{3}= \frac{4}{3} * 8 *a^{3}$

⇒ $a^{3} =8$

$a=2$

Hence, we can say that if the area of the region bounded by the curve $y^2 =4ax$ and the line $x= 4a$ is $\frac{256}{3}$ Sq units, then the value of $a$ will be $2$ .

To know more about Area bounded by the curve click here

brainly.com/question/13252576

#SPJ2

8 0

2 years ago

Read 2 more answers

Need the answer please and with work. I had put B but that is incorrect.

igor_vitrenko [27]

Answer:

Step-by-step explanation:

You answer goes to minus infinity. There is no answer and you are correct.

6 0

3 years ago

How many ninths does it take to make the same amount as 1/3.

Tems11 [23]

It takes 3/9 to make 1/3.

You multiply the 1 (in 1/3) by 3, and you multiply the 3 (in 1/3) by 3. You get 3/9. You needed to multiply by 3 to make 1/3 have a common denominator is 9.

7 0

4 years ago

Y=1/4 x 2^x<br> kkkkkkkk

levacccp [35]

Answer:

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Select the recursive rule that models the sequence below.14,20,26,32​

Select the recursive rule that models the sequence below.14,20,26,32