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3 years ago

10

How many cups of sorbet are used in a cups of punch

1 answer:

Ivanshal [37]3 years ago

7 0

How many cups of sorbet are used in 8 cups of punch?
<span> </span>
<span>15% * 8 cups </span>
<span>= 0.15 * 8 cups </span>
<span>= 1.2 cups </span>

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The circumference of the ellipse approximate. Which equation is the result of solving the formula of the circumference for b?

Serhud [2]

Answer:

$b = \sqrt{\frac{C^{2} }{2(\pi )^{2} } - a^{2}}$

Step-by-step explanation:

Given - The circumference of the ellipse approximated by $C = 2\pi \sqrt{\frac{a^{2} + b^{2} }{2} }$ where 2a and 2b are the lengths of 2 the axes of the ellipse.

To find - Which equation is the result of solving the formula of the circumference for b ?

Solution -

$C = 2\pi \sqrt{\frac{a^{2} + b^{2} }{2} }\\\frac{C}{2\pi } = \sqrt{\frac{a^{2} + b^{2} }{2} }$

Squaring Both sides, we get

$[\frac{C}{2\pi }]^{2} = [\sqrt{\frac{a^{2} + b^{2} }{2} }]^{2} \\\frac{C^{2} }{(2\pi)^{2} } = {\frac{a^{2} + b^{2} }{2} }\\2\frac{C^{2} }{4(\pi)^{2} } = {{a^{2} + b^{2} }$

$\frac{C^{2} }{2(\pi )^{2} } = a^{2} + b^{2} \\\frac{C^{2} }{2(\pi )^{2} } - a^{2} = b^{2} \\\sqrt{\frac{C^{2} }{2(\pi )^{2} } - a^{2}} = b$

∴ we get

$b = \sqrt{\frac{C^{2} }{2(\pi )^{2} } - a^{2}}$

8 0

3 years ago

How do you simplify <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B2%7D%20%2B%5Csqrt%7B6%7D%20%7D%7B%5Csqrt%7B8%7D%20%2B%

blondinia [14]

The trick is to exploit the difference of squares formula,

$a^2-b^2=(a-b)(a+b)$

Set a = √8 and b = √6, so that a + b is the expression in the denominator. Multiply by its conjugate a - b:

$(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)=(\sqrt8)^2-(\sqrt6)^2=8-6=2$

Whatever you do to the denominator, you have to do to the numerator too. So

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}{(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}2$

Expand the numerator:

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{2\cdot8}+\sqrt{6\cdot8}-\sqrt{2\cdot6}-(\sqrt6)^2$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{16}+\sqrt{48}-\sqrt{12}-6$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=4+\sqrt{48}-\sqrt{12}-6$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(\sqrt4-1)$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(2-1)$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6}=-2-\sqrt{12}$

So we have

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+\sqrt{12}}2$

But √12 = √(3•4) = 2√3, so

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+2\sqrt3}2=\boxed{-1-\sqrt3}$

7 0

2 years ago

(X-3)(x+2)=0 find the zeros

Ivanshal [37]

Answer: Min = (0.5, −6.25)

Step-by-step explanation: Standard form:

x2 − x − 6 = 0

Factorization:

(x + 2)(x − 3) = 0

Solutions based on factorization:

x + 2 = 0 ⇒ x1 = −2

x − 3 = 0 ⇒ x2 = 3

6 0

3 years ago

FIND THE area of the shaded figure

bezimeni [28]

Answer:

12

3 x 4

if its one unit then you just multiply the sides in this case 3 x 4

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Giving brainlyist and 5 stars

sweet-ann [11.9K]

Answer:

B: Each sandwich has one third pound of ham.

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