Right triangles 1, 2, and 3 are given with all their angle measures and approximate side lengths. Use one of the triangles to approximate the ratio (KL)/(JL).

- 0.64
- 0.77
- 0.83
- 1.2

Right Triangles 1, 2, And 3 Are Given With All Their Angle Measures And Approximate Side Lengths. Use
Right Triangles 1, 2, And 3 Are Given With All Their Angle Measures And Approximate Side Lengths. Use

Answers

Answer 1

Answer:

[tex]\frac{KL}{JL}[/tex] = [tex]\frac{6.4}{7.7}[/tex] = 0.83

Step-by-step explanation:

The key understanding here is that ΔJKL is similar to triangle 3 based on the AA criterion (they both have a right angle and a 40° angle).

We can find [tex]\frac{KL}{JL}[/tex] by setting up a proportion statement that includes KL, JL, and the lengths of their corresponding sides in triangle 3.

We can use this proportion:

[tex]\frac{KL}{6.4}[/tex] = [tex]\frac{JL}{7.7}[/tex]

[tex]\frac{KL}{6.4}[/tex] ⇒ opposite to 40° angle

KL, JL ⇒ ΔJKL

[tex]\frac{JL}{7.7}[/tex] ⇒ adjacent to 40° angle

6.4, 7.7 ⇒ triangle 3

[Now see the attachment]

Now we can rewrite the equation to show the ratios of the side lengths within each triangle.

[tex]\frac{KL}{JL}[/tex] = [tex]\frac{6.4}{7.7}[/tex]

[tex]\frac{KL}{JL}[/tex] ⇒ ΔJKL

KL, 6.4 ⇒ adjacent to 40° angle

[tex]\frac{6.4}{7.7}[/tex] ⇒ triangle 3

JL, 7.7 ⇒ adjacent to 40° angle

Right Triangles 1, 2, And 3 Are Given With All Their Angle Measures And Approximate Side Lengths. Use
Answer 2

Answer: it’s 0.82

Step-by-step explanation:


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Answers

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answer:

Step-by-step explanation:

You have four balls

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Step-by-step explanation:

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Step-by-step explanation:

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The function that shows this information is:

[tex]\large\boxed{c=4.20h}[/tex]

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Answers

Answer:

Step-by-step explanation:

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Step-by-step explanation:

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Answer:

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[tex]prove that\ \textless \ br /\ \textgreater \ \frac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \frac { 1 + \sin \theta } { \cos \theta }[/tex]
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0r any top user
who will answer correctly he will be a good user
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Answers

[tex] \large \bigstar \frak{ } \large\underline{\sf{Solution-}}[/tex]

Consider, LHS

[tex]\begin{gathered}\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } \\ \end{gathered}[/tex]

We know,

[tex]\begin{gathered}\boxed{\sf{  \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: \: }} \\ \end{gathered} 

\\ \\ \text{So, using this identity, we get}

\\ \\ \begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - ( {sec}^{2}\theta - {tan}^{2}\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}[/tex]

We know,

[tex]\begin{gathered}\boxed{\sf{  \:\rm \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: \: }} \\ \end{gathered}  \\ [/tex]

So, using this identity, we get

[tex]\begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - (sec\theta + tan\theta )(sec\theta - tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}[/tex]

can be rewritten as

[tex]\begin{gathered}\rm\:=\:\dfrac {(\sec \theta + tan\theta ) - (sec\theta + tan\theta )(sec\theta -tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac {(\sec \theta + tan\theta ) \: \cancel{(1 - sec\theta + tan\theta )}} { \cancel{ \tan \theta - \sec \theta + 1} } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:sec\theta + tan\theta \\\end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1}{cos\theta } + \dfrac{sin\theta }{cos\theta } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1 + sin\theta }{cos\theta } \\ \end{gathered}[/tex]

Hence,

[tex]\begin{gathered} \\ \rm\implies \:\boxed{\sf{  \:\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \:\dfrac{1 + sin\theta }{cos\theta } \: \: }} \\ \\ \end{gathered}[/tex]

[tex]\rule{190pt}{2pt}[/tex]

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Use KCF:
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F: Flip (flip your second faction numbers)

Ex.
Equation: 3/5 divided by 2/6
KCF: 3/5 multiplied by 6/2

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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A. a withdrawal or $232 followed by a deposit of $132

B. a deposit of $278 followed by a withdrawal of $278

C. a withdrawal of $115 followed by a deposit of $312

D. a deposit of $205 followed by a withdrawal of $317

Answers

The transactions that would leave the initial balance the same are a deposit of $278 followed by a withdrawal of $278.

The transactions that would leave the initial balance the same are transactions that net to zero. Withdrawals reduce the balance in the account. Deposits increase the value of the balance in the account.

Option A: Net balance

$275 - $232 + 132 = $175

Option B: Net balance

$275 + $278 - $278 = $275

Option C: Net balance

$275 - $115 + $312 = $472

Option D: Net balance

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Step-by-step explanation:

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Answer:

Base = Amount, and solve for the unknown numbers.Step-by-step explanation:

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Answers

The sample sizes are quite large, so the central limit theorem applies. (It typically does as soon as the sample size exceeds 30 or so.) This means that the sample mean will be approximately normally distributed with the same mean 22 but standard deviation 3.1/√40 ≈ 0.4902.

Now, if the question is asking about the probability of the sample mean being an exact number, that probability would be zero.

But if you meant to ask something else, like "what is the probability that the sample mean is less than 21?" then we would have a non-zero probability. In this particular case, if Y is a random variable for the sample mean, then

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