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3 years ago

Hans's Coffee Shop makes a blend that is a mixture of two types of coffee. Type A coffee costs Hans

1 answer:

3 0

He used 53 pounds. I got this by using two equations.
a+b=117
and
5.8a+4.1b= 569.8

I took the first equation and subtracted 'b' so the equation looked like a=117-b.
I then plugged in 117-b into the second equation, so it ended up looking like this:
5.8 (117-b)=569.8
5.8 times 117 is 678.6. And 5.8 times negative b is negative 5.8b.
The equation now looks like this:
678.6-5.8b+4.1b=569.8
Add like terms
678.6-1.7b=569.8
Then subtract 678.6 from both sides
-1.7b= -108.8
divide each side by -1.7
you get b= 64.
But because you are looking for how many pounds for a, subtract 64 from 117. You get 53. 53 is the answer.

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Answer:

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Step-by-step explanation:

(75, 60) (45, 40)

75 60 since 75 is larger than 45, we subtract = -30

45 40 since 60 is larger than 40, we subtract = -20

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2 years ago

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2 years ago

What is the exact value of sin(theta + beta)?

NISA [10]

Answer: $\frac{-3\sqrt{13}+4\sqrt{3}}{20}$

This is the single fraction of -3*sqrt(13)+4*sqrt(3) up top all over 20.

sqrt = square root

=======================================================

Explanation:

Angle theta is between pi and 3pi/2. This places the angle in quadrant Q3 where both cosine and sine are negative

Use the pythagorean trig identity to get the following:

$\sin^2 \theta + \cos^2 \theta = 1\\\\\sin^2 \theta + \left(-\frac{\sqrt{3}}{4}\right)^2 = 1\\\\\sin^2 \theta + \frac{3}{16} = 1\\\\\sin^2 \theta = 1 - \frac{3}{16}\\\\\sin^2 \theta = \frac{16}{16} - \frac{3}{16}\\\\\sin^2 \theta = \frac{16-3}{16}\\\\\sin^2 \theta = \frac{13}{16}\\\\\sin \theta = -\sqrt{\frac{13}{16}} \ \text{ ... sine is negative in Q3}\\\\\sin \theta = -\frac{\sqrt{13}}{\sqrt{16}}\\\\\sin \theta = -\frac{\sqrt{13}}{4}\\\\$

Angle beta is in Q1 where sine and cosine are positive.

Draw a right triangle with legs 3 and 4. The hypotenuse is 5 through the pythagorean theorem. In other words, we have a 3-4-5 right triangle.

Since $\tan \beta = \frac{3}{4}$ , this means $\sin \beta = \frac{3}{5} \ \text{ and } \ \cos \beta = \frac{4}{5}$

Use these ideas:

sin = opposite/hypotenuse
cos = adjacent/hypotenuse
tan = opposite/adjacent

In this case we have: opposite = 3, adjacent = 4, hypotenuse = 5.

-------------------------------------

To recap:

$\cos \theta = -\frac{\sqrt{3}}{4}\\\\\sin \theta = -\frac{\sqrt{13}}{4}\\\\\cos \beta = \frac{3}{5}\\\\\sin \beta = \frac{4}{5}\\\\$

They lead to this

$\sin\left(\theta + \beta\right) = \sin \theta * \cos \beta - \cos \theta * \sin \beta\\\\\sin\left(\theta + \beta\right) = -\frac{\sqrt{13}}{4} * \frac{3}{5} - \left(-\frac{\sqrt{3}}{4}\right) * \frac{4}{5}\\\\\sin\left(\theta + \beta\right) = -\frac{3\sqrt{13}}{20}+\frac{4\sqrt{3}}{20}\\\\\sin\left(\theta + \beta\right) = \frac{-3\sqrt{13}+4\sqrt{3}}{20}\\\\$

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2 years ago