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2 years ago

“solve each triangle to the nearest tenth of a unit”

Mathematics

1 answer:

Hitman42 [59]2 years ago

5 0

Answer:

Step-by-step explanation:Solve each triangle. Round side lengths to the

nearest tenth and angle measures to the nearest

degree.

SOLUTION:

Use the Law of Cosines to find the missing side

length.

Use the Law of Sines to find a missing angle

measure.

Find the measure of .

ANSWER:

A ≈ 36°, C ≈ 52°, b ≈ 5.1

SOLUTION:

Use the Law of Cosines to find the measure of the

largest angle, .

Use the Law of Sines to find3. a = 5, b = 8, c = 12

SOLUTION:

Use the Law of Cosines to find the measure of the

largest angle, .

Use the Law of Sines to find the measure of angle,

Find the measure of .

ANSWER:

A ≈ 18°, B ≈ 29°, C ≈ 133°

4. B = 110°, a = 6, c = 3

SOLUTION:

Use the Law of Cosines to find the missing side

length.

Use the Law of Sines to find a missing angle

measure.

Find the measure

ANSWER:

A ≈ 48°, C ≈ 22°, b ≈ 7.6

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In a linear Pair angles are?

tatiyna

Answer:

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5 0

3 years ago

Read 2 more answers

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

The school theater department made $2,142 on ticket sales for the three nights of their play. The department sold the same numbe

anyanavicka [17]

Answer:

102 tickets

Step-by-step explanation:

Let's say the total number of tickets sold is n.

7n = 2142

n = 2142 ÷ 7

n = 306

They sold 306 tickets in total.

We want to know how much they sold per night. We know they sold the same number of tickets each night for three nights.

306 ÷ 3 = 102

Therefore, they sold 102 tickets each night.

I hope this helps :)

6 0

3 years ago

Kylie explained that (negative 4 x + 9) squared will result in a difference of squares because (negative 4 x + 9) squared = (neg

Eva8 [605]

Given:

The given expression is:

$(-4x+9)^2$

According to Kylie,

$(-4x+9)^2=(-4x)^2+(9)^2$

$(-4x+9)^2=16x^2+81$

To find:

The correct statement for Kylie's explanation.

Solution:

We have,

$(-4x+9)^2$

According to the perfect square trinomial $(a+b)^2=a^2+2ab+b^2$ .

$(-4x+9)^2=(-4x)^2+2(-4x)(9)+(9)^2$

$(-4x+9)^2=16x^2-72x+81$

Kylie did not understand that this is a perfect square trinomial, and she did not determine the product correctly.

Therefore, the correct option is C.

4 0

3 years ago

Read 2 more answers

Ann works at a store in the mall and earns a wage of 8 dollars an hour. She earns 10 dollars an hour is she works on the weekend

soldier1979 [14.2K]

Answer:

The Answer is $352

Step-by-step explanation:

8 times 24 = 192

10 times 16 = 160

192 + 160 = 352

6 0

3 years ago