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3 years ago

12

Darrell earned a total of $10 by selling 8 cups of lemonade. How many cups of lemonade does darrell need to sell in all to earn

$25

2 answers:

ruslelena [56]3 years ago

8 0

Answer:

20

Step-by-step explanation:

each cup cost $1.25 . if he sells 12 more cups he'll have $25

goblinko [34]3 years ago

3 0

Answer:

Darrell will need to sell 20 cups of lemonade

Step-by-step explanation:

$10 = 8 cups of lemonade

8/8=1 10/8= 1.25

1 cup of lemonade costs $1.25

$25/$1.25=20

(/=division symbol)

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Brums [2.3K]

Answer:

B

Step-by-step explanation:

X∩Y is the intersection of X and Y which contains only the common elements

X∩Y = { }

Since there are no common elements

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4 years ago

Solve Sine inverse x +sine inverse 2x = π/3

sergeinik [125]

$\arcsin x+\arcsin2x=\dfrac\pi3$
$\sin\left(\arcsin x+\arcsin2x\right)=\sin\dfrac\pi3$
$\sin(\arcsin x)\cos(\arcsin2x)+\sin(\arcsin2x)\cos(\arcsin x)=\dfrac{\sqrt3}2$

Note that $\arcsin x$ is defined for $|x|\le1$ , and $\arcsin2x$ is defined for $|x|\le\dfrac12$ , where the latter will be the "total" domain. Under this condition, you have $\sin(\arcsin x)=x$ and $\sin(\arcsin2x)=2x$ . The cosine terms can be found with Pythagoras' theorem.

So provided that $|x|\le\dfrac12$ , it follows that the above reduces to

$x\sqrt{1-4x^2}+2x\sqrt{1-x^2}=\dfrac{\sqrt3}2$

Squaring both sides gives

$x^2(1-4x^2)+4x^2\sqrt{(1-4x^2)(1-x^2)}+4x^2(1-x^2)=\dfrac34$
$5x^2-8x^4+4x^2\sqrt{(1-4x^2)(1-x^2)}=\dfrac34$
$8x^4-5x^2+\dfrac34=4x^2\sqrt{(1-4x^2)(1-x^2)}$

Squaring both sides again gives

$\left(8x^4-5x^2+\dfrac34\right)^2=16x^4(1-4x^2)(1-x^2)$
$64x^8-80x^6+37x^4-\dfrac{15}2x^2+\dfrac9{16}=64x^8-80x^6+16x^4$
$21x^4-\dfrac{15}2x^2+\dfrac9{16}=0$
$112x^4-40x^2+3=0$
$(4x^2-1)(28x^2-3)=0$
$(2x-1)(2x+1)(\sqrt{28}x-\sqrt3)(\sqrt{28}x+\sqrt3)=0$

$\implies x=\pm\dfrac12,\pm\sqrt{\dfrac3{28}}$

Three of these solutions are extraneous, however.

When $x=-\dfrac12$ , we have $\arcsin x+\arcsin2x=-\dfrac{2\pi}3$ .

When $x=\dfrac12$ , we have $\arcsin x+\arcsin2x=\dfrac{2\pi}3$ .

When $x=-\sqrt{\dfrac3{28}}$ , we have $\arcsin x+\arcsin2x=-\dfrac\pi3$ .

Finally, when $x=\sqrt{\dfrac3{28}}$ , we have $\arcsin x+\arcsin2x=\dfrac\pi3$ , so this is our only solution.

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3 years ago

A sound wave is modeled with the equation y = 1 4 cos 2 π 3 θ . a. Find the period. Explain your method. b. Find the amplitude.

Fittoniya [83]

If the frequency of the sound wave is 2π/3, then the time period of the sound wave will be 0.447 seconds. Then the amplitude is 14.

<h3>What is wavelength?</h3>

The distance between two successive troughs or crests is known as the wavelength. The peak of the wave is the highest point, while the trough is the lowest.

The wavelength is also defined as the distance between two locations in a wave that have the same oscillation phase.

A sound wave is modeled with the equation y = 14 cos (2π/3) θ.

The frequency of the sound wave will be

$f = \dfrac{2\pi}{3}$

Then the time period will be

$T = \dfrac{1}{f}\\\\T = \dfrac{1}{2\pi /3}\\\\T = \dfrac{3}{2\pi}\\\\T = 0.477 \ \rm seconds$

Then the amplitude of the sound wave will be

A = 14

To learn more about the wavelength refer to the link;

brainly.com/question/7143261

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2 years ago

Which of the following are true statements about angle parking. Angle parking spots have a larger blind spaces than perpendicula

S_A_V [24]

Answer:

Angle parking is more common than perpendicular parking.

Angle parking spots have half the blind spot as compared to perpendicular parking spaces

Step-by-step explanation:

Considering the available options, the true statement about angle parking is that" Angle parking is more common than perpendicular parking." Angle parking is mostly constructed and used for public parking. It is mostly used where the parking lots are quite busy such as motels or public garages.

Therefore, in this case, the answer is that "Angle parking is more common than perpendicular parking."

Also, "Angle parking spots have half the blind spot as compared to perpendicular parking spaces."

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3 years ago

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I think it’s C but what grade is this

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