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e-lub [12.9K]
3 years ago
12

Jovan made batches of 12 fruit bars with fruit bars in each batch. He kept 2 batches of fruit bars at home for his family and br

ought the rest to a school bake sale. Saleh also brought fruit bars for the bake sale. Combined, they had 200 fruit bars for the bake sale. Which expression represents the number of fruit bars Saleh brought to the bake sale? help please and Thank you

Mathematics
1 answer:
Galina-37 [17]3 years ago
5 0
I think it’s A 200-12(b-2)
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What is the completely factored form of d4 -8d2 + 16
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According to Vieta's Formulas, if x_1,x_2 are solutions of a given quadratic equation:

ax^2+bx+c=0

Then:

a(x-x_1)(x-x_2) is the completely factored form of ax^2+bx+c.

If choose x=d^2, then:

\displaystyle x^2-8x+16=0\\\\x_{1,2}= \frac{8\pm  \sqrt{64-64} }{2}=4

So, according to Vieta's formula, we can get:

x^2-8x+16=(x-4)(x-4)= (x-4)^2

But x=d^2:

d^4-8d^2+16=(d^2-4)^2=[(d+2)(d-2)]^2=(d+2)^2(d-2)^2
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3 years ago
Sandy makes linen scarves that are 7/8 of a yard long. How many scarves can she make from 156 feet of fabric?
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4 years ago
In a class there are 15 students. 8 of them like playing soccer, 6 of them like swimming, and 2 like both and swimming and playi
xenn [34]

Answer:

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Marco and Drew stacked boxes on a shelf. Marco lifted 9 boxes and Drew lifted 14 boxes. The boxes that Drew lifted each weighed
belka [17]
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Plźzzz help me ... I'll give brainliest
fgiga [73]

Answer:

\displaystyle x = \frac{1}{3}.

Step-by-step explanation:

Consider the double angle identity for tangents:

\displaystyle \tan(2\theta) = \frac{2\tan{\theta}}{1 - \tan^{2}{\theta}}.

Also, for two complementary angles,

\displaystyle \tan{\left(\frac{\pi}{2} - \theta \right)} = \frac{1}{\tan{\theta}}.

Subtract \tan^{-1}(x + 1) from both sides of this equation:

\displaystyle 2\tan^{-1}{x} = \frac{\pi}{2} + (-\tan^{-1}(x + 1)).

Take the tangent of both sides of this equation:

\displaystyle \tan(2\tan^{-1}{x}) = \tan\left(\frac{\pi}{2} -\tan^{-1}\left(x + 1\right)\right).

Apply the double-angle identity to the left-hand side of this equation:

\displaystyle \tan(2\tan^{-1}{x}) \implies \frac{2\tan(\tan^{-1}x)}{1 - \tan^{2}(\tan^{-1}x)}\implies \frac{2x}{1 - x^{2}}.

The two angles \displaystyle \left(\frac{\pi}{2} + (-\tan^{-1}\left(x + 1\right))\right) and (x - 1) are complementary. Therefore, for the right-hand side of this equation,

\displaystyle \tan\left(\frac{\pi}{2} -\tan^{-1}\left(x + 1\right)\right)\implies \frac{1}{\tan(\tan^{-1}(x + 1))} \implies \frac{1}{x + 1}.

Equate the two sides of this equation:

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5 0
3 years ago
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