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2 years ago

The following two right triangles are similar.If side PQ = 42, side LM = 28, and side QR = 24, what is the length of side MN?

Mathematics

1 answer:

Phantasy [73]2 years ago

6 0

Answer:

a. 16

Step-by-step explanation:

Find the scale:

42 ÷ 28 = 1.5

Use this scale for QR to find MN:

24 ÷ 1.5 = 16

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The minute hand on a clock is 9 centimeters long and travels through an arc of 252 degrees every 42 minutes. To the nearest tent

kkurt [141]

Answer:

39.6 cm

Step-by-step explanation:

Applying

s = 2πrθ/360................ Equation 1

Where s = length of an arc or distance traveled by the minutes hand of the clock during the 42 munites, r = length of the minutes hand of the clock, θ = Angle traveled by the minute hand of the clock for every 42 minutes

From the question,

Given: r = 9 cm, θ = 252°

Constant: π = 22/7 = 3.14

Substitute these values into equation 1

s = (2×3.14×9×252)/360

s = 39.564

s = 39.6 cm

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2 years ago

Carlos and Maria drove a total of 228 miles in 4.6 hours. Carlos drove the first part of the trip and averaged

aleksklad [387]

Answer:

around 228

Step-by-step explanation:

8 0

2 years ago

PLEASE HELP!!!!!

Katena32 [7]

Answer:6018

Step-by-step explanation:

Given Sequence

$-282-266-250-234.....$

It represent an A.P. with

first term $a=-282$

common difference $d=16$

So sum of 51 term

$S_n=\frac{n}{2}[2a+(n-1)d]$

$S_{51}=\frac{51}{2}\times [2\times (-282)+(51-1)16]$

$S_{51}=\frac{51}{2}\times [-564+800]$

$S_{51}=\frac{51}{2}\times [236]$

$S_{51}=51\times 118$

$S_{51}=6018$

7 0

2 years ago

Justin now has a final exam today. There are 20 questions with 3 options. Each

mamaluj [8]

Answer: .0142%

Step-by-step explanation:

6 0

2 years ago

Prove that tan 10+tan 70'+taniou = tanio tanfo'tam 1oo.

Galina-37 [17]

Question:

Prove that:

$tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)$

Answer:

Proved

Step-by-step explanation:

Given

$tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)$

Required

Prove

$tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)$

Subtract tan(10) from both sides

$- tan(10)+tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100) - tan(10)$

$tan(70) + tan(100) = tan(10). tan(70). tan(100) - tan(10)$

Factorize the right hand size

$tan(70) + tan(100) = -tan(10)(-tan(70). tan(100) + 1)$

Rewrite as:

$tan(70) + tan(100) = -tan(10)(1-tan(70). tan(100))$

Divide both sides by $1-tan(70). tan(100)$

$\frac{tan(70) + tan(100)}{1-tan(70). tan(100)} = \frac{-tan(10)(1-tan(70). tan(100))}{1-tan(70). tan(100))}$

$\frac{tan(70) + tan(100)}{1-tan(70). tan(100)} = -tan(10)$

In trigonometry:

$tan(A + B) = \frac{tan(A) + tan(B)}{1 - tan(A)tan(B)}$

So:

$\frac{tan(70) + tan(100)}{1 - tan(70)tan(100)}$ can be expressed as: $tan(70 + 100)$

$\frac{tan(70) + tan(100)}{1-tan(70). tan(100)} = -tan(10)$ gives

$tan(70 + 100) = -tan(10)$

$tan(170) = -tan(10)$

In trigonometry:

$tan(180 - \theta) = -tan(\theta)$

So:

$tan(180 - 10) = -tan(10)$

Because RHS = LHS

Then:

$tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)$ has been proven

6 0

2 years ago