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3 years ago

Write an expression for the apparent nth term(a) of the sequence. 2,5,8,11,14...

Mathematics

1 answer:

ExtremeBDS [4]3 years ago

7 0

Answer:

Step-by-step explanation:

Since the sequence is increasing by a constant factor ( $3$ ) each time, we know that this is an arithmetic sequence.

The formula for an arithmetic sequence is the following:

$a_{n} = a_{1} + (n - 1)d$

Where $n$ is the number in the sequence and $d$ is the constant that the sequence is increasing by.

Based on all of the information given, we know that $a_{1} = 2$ (since $2$ is the first number in the sequence), and $d = 3$ , so we can construct the formula:

$a_{n} = 2 + (n - 1)3$

$a_{n} = 2 + 3n - 3$

$a_{n} = 3n - 1$

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Please help!! What is the solution to the quadratic inequality? 6x2≥10+11x

fredd [130]

Answer:

The solution of the inequation $6\cdot x^{2} \geq 10 + 11\cdot x$ is $\left(-\infty,-\frac{2}{3}\right]\cup\left[\frac{5}{2},+\infty\right)$ .

Step-by-step explanation:

First of all, let simplify and factorize the resulting polynomial:

$6\cdot x^{2} \geq 10 + 11\cdot x$

$6\cdot x^{2}-11\cdot x -10 \geq 0$

$6\cdot \left(x^{2}-\frac{11}{6}\cdot x -\frac{10}{6} \right)\geq 0$

Roots are found by Quadratic Formula:

$r_{1,2} = \frac{\left[-\left(-\frac{11}{6}\right)\pm \sqrt{\left(-\frac{11}{6} \right)^{2}-4\cdot (1)\cdot \left(-\frac{10}{6} \right)} \right]}{2\cdot (1)}$

$r_{1} = \frac{5}{2}$ and $r_{2} = -\frac{2}{3}$

Then, the factorized form of the inequation is:

$6\cdot \left(x-\frac{5}{2}\right)\cdot \left(x+\frac{2}{3} \right)\geq 0$

By Real Algebra, there are two condition that fulfill the inequation:

a) $x-\frac{5}{2} \geq 0 \,\wedge\,x+\frac{2}{3}\geq 0$

$x \geq \frac{5}{2}\,\wedge\,x \geq-\frac{2}{3}$

$x \geq \frac{5}{2}$

b) $x-\frac{5}{2} \leq 0 \,\wedge\,x+\frac{2}{3}\leq 0$

$x \leq \frac{5}{2}\,\wedge\,x\leq-\frac{2}{3}$

$x\leq -\frac{2}{3}$

The solution of the inequation $6\cdot x^{2} \geq 10 + 11\cdot x$ is $\left(-\infty,-\frac{2}{3}\right]\cup\left[\frac{5}{2},+\infty\right)$ .

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3 years ago

Find the measure indicated. Assume that the lines which appear to be tangent are tangent. PART 3

ludmilkaskok [199]

I'll do problems 13 and 14 to get you started.

=====================================================

Problem 13

<h3>Answer: 32 degrees</h3>

-----------------------

Explanation:

Refer to the diagram below for problem 13. Since BD is a tangent of the circle, this means angle ABD is 90 degrees, and furthermore x+y = 90. This solves to y = 90-x.

If we focus on triangle ABC, then we can add the interior angles and set the sum equal to 180

A+B+C = 180

z+y+y = 180

z+2y = 180

z+2(90-x) = 180 .... plug in y = 90-x

z+180-2x = 180

z-2x = 180-180

z-2x = 0

z = 2x

We'll use this later.

Now move onto triangle ABD. This is a right triangle due to angle ABD being 90 degrees. The acute angles z and 26 are complementary, meaning they add to 90. So,

z+26 = 90

z = 90-26

z = 64

Now plug this into z = 2x and solve for x

z = 2x

64 = 2x

2x = 64

x = 64/2

x = 32

=====================================================

Problem 14

<h3>Answer: 31 degrees</h3>

-----------------------

Explanation:

Refer to the image shown below for problem 14. Like before, I've added various labels to it to help find the angle we're after (the red angle x).

Note how I've drawn a dashed line from point A to point D. This splits up quadrilateral ABCD into two triangles ABD and ACD. These triangles can be shown to be congruent using the HL (hypotenuse leg) theorem.

The congruent triangles also implies that angle DBC and angle BCD are the same measure (both are shown in green as the variable y).

Focus on triangle BCD. It has interior angles of: y, y and 62. Add them up, set the sum equal to 180.

y+y+62 = 180

2y+62 = 180

2y = 180-62

2y = 118

y = 118/2

y = 59

As the image attachment mentions, the angles x and y add to 90 degrees. This is because angle ABD is 90 degrees, due to segment BD being tangent to the circle. So x+y = 90 leads to x = 90-y, and therefore,

x = 90-y

x = 90-59

x = 31

Side note: it is not a coincidence that we ended up with half of the value of 62.