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2 years ago

Certain flightness beetles havw wing

Mathematics

1 answer:

Doss [256]2 years ago

3 0

Answer:

yes. but they cant fly cuz the wings are jus for show or too scare predaters

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What is the probability that a point chosen at random in the given figure will be inside the larger circle and outside the small

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<span>The probability that a point is chosen at random in the given figure is 40/49 or .82. The answer to the following statement or the missing blank is 40/49. The probability is the possibility of the occurrence or happening of something.</span>

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3 years ago

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a baseball team paid eight times as much to their ace pitcher as they paid to their shortstop. if they paid 81 million for the t

GenaCL600 [577]

x +y = 81 million

let x be the short stop and y be the pitcher

so pitcher (y) gets 2 times what the the shortstop (x) gets so y = 2x

so x + 2x = 81 million

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3 years ago

Solve 6 sin((π/5)x)=5 for the four smallest positive solutions

cupoosta [38]

Answer:

1.568, 3.432, 11.568, 13.432

Step-by-step explanation:

Divide equation $6\sin\left(\dfrac{\pi }{5}x\right)=5$ by 6:

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Then

$\dfrac{\pi }{5}x=(-1)^k\arcsin\left(\dfrac{5}{6}\right)+\pi k,\ k\in Z,$

$x=(-1)^k\dfrac{5}{\pi }\arcsin\left(\dfrac{5}{6}\right)+5k,\ k\in Z.$

Since $\arcsin\left(\dfrac{5}{6}\right)\approx 56^{\circ}\approx \dfrac{\pi }{3.2},$

four smallest positive solutions are

1. $\dfrac{5}{3.2}\approx 1.568,\ k=0;$

2. $\dfrac{-5}{3.2}+5\approx 3.432,\ k=1;$

3. $\dfrac{5}{3.2}+10\approx 11.568,\ k=2;$

4. $\dfrac{-5}{3.2}+15\approx 13.432,\ k=3.$

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3 years ago

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AlladinOne [14]

First subtract 9/4 from each side

1/2b=-2/4

Simplify the 2/4

1/2b=-1/2

Divide each side by 1/2

b = -1

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3 years ago

Write an expression for the sequence of operations described below.

Aleks [24]

Answer:

g/(2 x 3)

Step-by-step explanation:

Your expression will be:

-Double 3 = multiply 3 by 2

-Divide g by the result (2 × 3)

g/(2 × 3)

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3 years ago

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Certain flightness beetles havw wing​

Certain flightness beetles havw wing