274,635 rounded to the nearest thousands

In the following figure, the value of x is____

zubka84 [21]

Answer:

x ≈ 8.3 ( to the nearest tenth )

Step-by-step explanation:

Using the cosine ratio in the right triangle

cos41° = $\frac{adjacent}{hypotenuse}$ = $\frac{BC}{AB}$ = $\frac{x}{11}$

Multiply both sides by 11

11 × cos41° = x, thus

x ≈ 8.3

7 0

3 years ago

A. Write two explicit formulas for arithmetic sequences.

IgorLugansk [536]

Explicit formulas for arithmetic sequences are derived from terms in arithmetic sequences. It helps to find each term in arithmetic progression easily. The arithmetic progression is a1, a2, a3, ..., an. where the first term is denoted as 'a', we have a = a1, and the tolerance is denoted as 'd'. The tolerance formula is d = a2 - a1 = a3 - a2 = an - an - 1. The nth term of the arithmetic progression represents the explicit formula for the arithmetic progression.

Explicit formula: an= a + (n − 1) d

Explicit formula: Sn = n/2 [2a+(n-1) d]

Where,

nth term in the arithmetic sequence

a = first term in the arithmetic sequence

d = difference (each term and its term difference) previous term, i.e., d = an-an-1

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7 0

1 year ago

This year the winner of New York Marathon ran the 8 kilometer event in 24 hours. What is the runner’s unit rate?

ValentinkaMS [17]

3 because 24 divide by 8 is 3

8 0

3 years ago

PLEASE ANSWERRRRRRRRR

chubhunter [2.5K]

Answer:

0

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$