All categories

3 years ago

Helpppppppppppppppppppp

Mathematics

1 answer:

Sergio [31]3 years ago

6 0

Answer:

what helpppppppp???

Step-by-step explanation:

????

You might be interested in

The measure of an angle is 78°. What is the measure of a complementary angle?

ki77a [65]

First you need to know the definition of a complementary angle. That means that two angles add up to 90 degrees, or a right angle.

To find an angle complementary to a 78 degree angle, you need to subtract 78 from 90.

90 - 78 = 12

6 0

3 years ago

Read 2 more answers

The basketball team at Martin Luther King High School won 12 of its first 15 games. Then it won its next 3 games. The percent of

aliya0001 [1]

12/15 games have been won then they won 3 which makes it then 15/18 because the denominator is how many games they've played while the numerator is how many games they one.

$\frac{15}{18} = \frac{x}{100}$
Cross multiply both...
$1500=18x$
divide each side by 18 to make the x by itself
Which gives you x=83.333%

4 0

3 years ago

2,17,82,257,626,1297 next one please ?

In-s [12.5K]

The easy thing to do is notice that 1^4 = 1, 2^4 = 16, 3^4 = 81, and so on, so the sequence follows the rule $n^4+1$ . The next number would then be fourth power of 7 plus 1, or 2402.

And the harder way: Denote the n-th term in this sequence by $a_n$ , and denote the given sequence by $\{a_n\}_{n\ge1}$ .

Let $b_n$ denote the n-th term in the sequence of forward differences of $\{a_n\}$ , defined by

$b_n=a_{n+1}-a_n$

for n ≥ 1. That is, $\{b_n\}$ is the sequence with

$b_1=a_2-a_1=17-2=15$

$b_2=a_3-a_2=82-17=65$

$b_3=a_4-a_3=175$

$b_4=a_5-a_4=369$

$b_5=a_6-a_5=671$

and so on.

Next, let $c_n$ denote the n-th term of the differences of $\{b_n\}$ , i.e. for n ≥ 1,

$c_n=b_{n+1}-b_n$

so that

$c_1=b_2-b_1=65-15=50$

$c_2=110$

$c_3=194$

$c_4=302$

etc.

Again: let $d_n$ denote the n-th difference of $\{c_n\}$ :

$d_n=c_{n+1}-c_n$

$d_1=c_2-c_1=60$

$d_2=84$

$d_3=108$

etc.

One more time: let $e_n$ denote the n-th difference of $\{d_n\}$ :

$e_n=d_{n+1}-d_n$

$e_1=d_2-d_1=24$

$e_2=24$

etc.

The fact that these last differences are constant is a good sign that $e_n=24$ for all n ≥ 1. Assuming this, we would see that $\{d_n\}$ is an arithmetic sequence given recursively by

$\begin{cases}d_1=60\\d_{n+1}=d_n+24&\text{for }n>1\end{cases}$

and we can easily find the explicit rule:

$d_2=d_1+24$

$d_3=d_2+24=d_1+24\cdot2$

$d_4=d_3+24=d_1+24\cdot3$

and so on, up to

$d_n=d_1+24(n-1)$

$d_n=24n+36$

Use the same strategy to find a closed form for $\{c_n\}$ , then for $\{b_n\}$ , and finally $\{a_n\}$ .

$\begin{cases}c_1=50\\c_{n+1}=c_n+24n+36&\text{for }n>1\end{cases}$

$c_2=c_1+24\cdot1+36$

$c_3=c_2+24\cdot2+36=c_1+24(1+2)+36\cdot2$

$c_4=c_3+24\cdot3+36=c_1+24(1+2+3)+36\cdot3$

and so on, up to

$c_n=c_1+24(1+2+3+\cdots+(n-1))+36(n-1)$

Recall the formula for the sum of consecutive integers:

$1+2+3+\cdots+n=\displaystyle\sum_{k=1}^nk=\frac{n(n+1)}2$

$\implies c_n=c_1+\dfrac{24(n-1)n}2+36(n-1)$

$\implies c_n=12n^2+24n+14$

$\begin{cases}b_1=15\\b_{n+1}=b_n+12n^2+24n+14&\text{for }n>1\end{cases}$

$b_2=b_1+12\cdot1^2+24\cdot1+14$

$b_3=b_2+12\cdot2^2+24\cdot2+14=b_1+12(1^2+2^2)+24(1+2)+14\cdot2$

$b_4=b_3+12\cdot3^2+24\cdot3+14=b_1+12(1^2+2^2+3^2)+24(1+2+3)+14\cdot3$

and so on, up to

$b_n=b_1+12(1^2+2^2+3^2+\cdots+(n-1)^2)+24(1+2+3+\cdots+(n-1))+14(n-1)$

Recall the formula for the sum of squares of consecutive integers:

$1^2+2^2+3^2+\cdots+n^2=\displaystyle\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6$

$\implies b_n=15+\dfrac{12(n-1)n(2(n-1)+1)}6+\dfrac{24(n-1)n}2+14(n-1)$

$\implies b_n=4n^3+6n^2+4n+1$

$\begin{cases}a_1=2\\a_{n+1}=a_n+4n^3+6n^2+4n+1&\text{for }n>1\end{cases}$

$a_2=a_1+4\cdot1^3+6\cdot1^2+4\cdot1+1$

$a_3=a_2+4(1^3+2^3)+6(1^2+2^2)+4(1+2)+1\cdot2$

$a_4=a_3+4(1^3+2^3+3^3)+6(1^2+2^2+3^2)+4(1+2+3)+1\cdot3$

$\implies a_n=a_1+4\displaystyle\sum_{k=1}^3k^3+6\sum_{k=1}^3k^2+4\sum_{k=1}^3k+\sum_{k=1}^{n-1}1$

$\displaystyle\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}4$

$\implies a_n=2+\dfrac{4(n-1)^2n^2}4+\dfrac{6(n-1)n(2n)}6+\dfrac{4(n-1)n}2+(n-1)$

$\implies a_n=n^4+1$

4 0

3 years ago

Two sides of a parallelogram measure 60 centimeters and 40 centimeters. If one angle of the parallelogram measures 1320, find th

UkoKoshka [18]

Answer:

b. about 91.7 cm and 44.6 cm

Step-by-step explanation:

The lengths of the diagonals can be found using the Law of Cosines.

Consider the triangle(s) formed by a diagonal. The two given sides will form the other two sides of the triangle, and the corner angles of the parallelogram will be the measure of the angle between those sides (opposite the diagonal).

For diagonal "d" and sides "a" and "b" and corner angle D, we have ...

d² = a² +b² -2ab·cos(D)

The measure of angle D will either be the given 132°, or the supplement of that, 48°. We can use the fact that the cosines of an angle and its supplement are opposites. This means the diagonal measures will be ...

d² = 60² +40² -2·60·40·cos(D) ≈ 5200 ±4800(0.66913)

d² ≈ {1988.2, 8411.8}

d ≈ {44.6, 91.7} . . . . centimeters

The diagonals are about 91.7 cm and 44.6 cm.

4 0

3 years ago

How to Evaluate 5/9 x 8

GuDViN [60]

Answer:

4 4/9

Step-by-step explanation:

First, you have to set it up: 5/9 times 8/1

We made it 8/1 since it has to be a fraction in the multipying ocess and the one does not change the value.

Next, Multiply.

5/9x8/1=40/9 as an inproper fraction

As a proper fraction, it is written as 4 4/9

3 0

2 years ago