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3 years ago

A triangle has sides with lengths of 100 inches, 56 inches, and 80 inches. Is it a right triangle?

Mathematics

2 answers:

Kryger [21]3 years ago

8 0

Answer:

No.

Step-by-step explanation:

If these sides come out true in the Pythagorean Theorem, then it is a right triangle.

a² + b² = c²

56² + 80² = 100²

3136 + 6400 = 10000

9536 ≠ 10000

Since these values do not come out true in the Pythagorean Theorem, it is not a right triangle.

weeeeeb [17]3 years ago

8 0

Answer:

Possibly, A right triangle has a really long side that is "c" in the picture. If the 100 length is c then the 80 one could be "b" while the 56 is "a"

I know it doesnt make much sense but I hope it helps

I see people looking at mine because its simple but you should look at the other guys awnser. He has information and details that explain their awnser. I didn't really do much to get the awnser. Look at theirs Please.

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Will report anybody who gives a link! 3,947 ÷ 45 = (?) R (?)

Free_Kalibri [48]

Answer:

87.7(1)

87 Remainder 32

8 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

Jackie plans to place a rectangular piece of art inside a rectangular frame. She considers a piece of art that is 0.5 meters lon

tamaranim1 [39]

Answer:

The art is rectangular because $(0.5) ^2 + (1.2) ^2 = (1.3) ^2$

(0.5) squared + (1.2) squared = (1.3) squared

Step-by-step explanation:

Given that the piece of art is rectangular in shape.

Length of piece of art = 0.5 meters

Width of piece of art = 1.2 meters

Kindly refer to the attached image in the answer area.

Two adjacent sides of a rectangle are given, the diagonal value can be found by using Pythagorean Theorem.

According to Pythagorean theorem:

$\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}$

Here, Hypotenuse will be the diagonal of the rectangle.

Base and Perpendicular will be the two adjacent sides.

Therefore,

$Diagonal^2 = 0.5^2+0.12^2\\\Rightarrow Diagonal^2 = 0.25+1.44\\\Rightarrow Diagonal = \sqrt{1.69}\\\Rightarrow Diagonal = 1.3\ m$

Therefore, the answer is:

The art is rectangular because $(0.5) ^2 + (1.2) ^2 = (1.3) ^2$

(0.5) squared + (1.2) squared = (1.3) squared

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