All categories

3 years ago

Find a formula for an arithmetic sequence that begins 80, 83, 86, 89, ...

Mathematics

1 answer:

mel-nik [20]3 years ago

3 0

Answer:

$a_{n}$ = 3n + 77

Step-by-step explanation:

The nth term of an arithmetic sequence is

$a_{n}$ = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = 80 and d = a₂ - a₁ = 83 - 80 = 3 , then

$a_{n}$ = 80 + 3(n - 1) = 80 + 3n - 3 = 3n + 77

You might be interested in

Look at the picture

Fantom [35]

Answer:

The x-intercept is 2.

Step-by-step explanation:

$2 \sqrt[3]{x - 10} + 4 = 0$

$2 \sqrt[3]{x - 10} = - 4$

$\sqrt[3]{x - 10 } = - 2$

$x - 10 = - 8$

$x = 2$

3 0

1 year ago

Sin 75°(cot 75° + cot 60°) help me simplify this

leva [86]

(square root of 6)/ 3

7 0

2 years ago

Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

iragen [17]

Answer:

36.25

Step-by-step explanation:

The three in 20.342 has a value of 0.3, so a three one hundred times greater would be 0.3 times 100, which is 30.

Next, find the choice that has a three in the ten's place, which represents a 30.

5 0

2 years ago

The sum of three consecutive integer is -204. what is the largest integer? -63 --44--67 -62

Lisa [10]

-x + -x-1 + -x-2 = -204
-3x=-201
x=-67

-67-68-69=-204

5 0

3 years ago

Use the discriminant to determine the number of solutions to the quadratic equation 3x^2+5x=-1

kari74 [83]

Answer:

Two real distinct solutions

Step-by-step explanation:

Hi there!

Background of the Discriminant

The discriminant $b^2-4ac$ applies to quadratic equations when they are organised in standard form: $ax^2+bx+c=0$ .

All quadratic equations can be solved with the quadratic formula: $x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}}$ .

When $b^2-4ac$ is positive, it is possible to take its square root and end up with two real, distinct values of x.

When it is zero, we won't be taking the square root at all and we will end up with two real solutions that are equal, or just one solution.

When it is negative, it is impossible to take the square root and we will end up with two non-real solutions.

Solving the Problem

$3x^2+5x=-1$

We're given the above equation. It hasn't been organised completely in $ax^2+bx+c=0$ , but we can change that by adding 1 to both sides to make the right side equal to 0:

$3x^2+5x+1=0$

Now that we can identify the values of a, b and c, we can plug them into the discriminant:

$D=b^2-4ac\\D=(5)^2-4(3)(1)\\D=25-4(3)(1)\\D=25-12\\D=13$

Therefore, because the discriminant is positive, the equation has two real, distinct solutions.

I hope this helps!

4 0

3 years ago

Find a formula for an arithmetic sequence that begins 80, 83, 86, 89, ...​

Find a formula for an arithmetic sequence that begins 80, 83, 86, 89, ...