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3 years ago

Solve the equation –15 = –6(3 + x) + 4(x – 6). X =

Mathematics

1 answer:

madam [21]3 years ago

7 0

Answer:

x=-13.5

Step-by-step explanation:

$-15=-6(3+x)+4(x-6)\\-15=-18-6x+4x-24\\-15=-2x-42\\27=-2x\\x=\frac{-27}{2}$

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What is the letter for these 2

Svetach [21]

1. B (opposite/hypotenuse)
2. C (opposite/adjacent)

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3 years ago

Read 2 more answers

I’ll brainliest you please help me

Tanya [424]

X= 75
y= 105
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3 years ago

Use calculus to find the largest possible area for a rectangular field that can be enclosed with a fence that is 400 meters long

Setler79 [48]

the largest rectangular area is actually a square.

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3 years ago

Need help with trig problem in pic

Sidana [21]

Answer:

a) $cos(\alpha)=-\frac{3}{5}\\$

b) $\sin(\beta)= \frac{\sqrt{3} }{2}$

c) $\frac{4+3\sqrt{3} }{10}\\$

d) $\alpha\approx 53.1^o$

Step-by-step explanation:

a) The problem tells us that angle $\alpha$ is in the second quadrant. We know that in that quadrant the cosine is negative.

We can use the Pythagorean identity:

$tan^2(\alpha)+1=sec^2(\alpha)\\(-\frac{4}{3})^2 +1=sec^2(\alpha)\\sec^2(\alpha)=\frac{16}{9} +1\\sec^2(\alpha)=\frac{25}{9} \\sec(\alpha) =+/- \frac{5}{3}\\cos(\alpha)=+/- \frac{3}{5}$

Where we have used that the secant of an angle is the reciprocal of the cos of the angle.

Since we know that the cosine must be negative because the angle is in the second quadrant, then we take the negative answer:

$cos(\alpha)=-\frac{3}{5}$

b) This angle is in the first quadrant (where the sine function is positive. They give us the value of the cosine of the angle, so we can use the Pythagorean identity to find the value of the sine of that angle:

$cos (\beta)=\frac{1}{2} \\\\sin^2(\beta)=1-cos^2(\beta)\\sin^2(\beta)=1-\frac{1}{4} \\\\sin^2(\beta)=\frac{3}{4} \\sin(\beta)=+/- \frac{\sqrt{3} }{2} \\sin(\beta)= \frac{\sqrt{3} }{2}$

where we took the positive value, since we know that the angle is in the first quadrant.

c) We can now find $sin(\alpha -\beta)$ by using the identity:

$sin(\alpha -\beta)=sin(\alpha)\,cos(\beta)-cos(\alpha)\,sin(\beta)\\$

Notice that we need to find $sin(\alpha)$ , which we do via the Pythagorean identity and knowing the value of the cosine found in part a) above:

$sin(\alpha)=\sqrt{1-cos^2(\alpha)} \\sin(\alpha)=\sqrt{1-\frac{9}{25} )} \\sin(\alpha)=\sqrt{\frac{16}{25} )} \\sin(\alpha)=\frac{4}{5}$

Then:

$sin(\alpha -\beta)=\frac{4}{5}\,\frac{1}{2} -(-\frac{3}{5}) \,\frac{\sqrt{3} }{2} \\sin(\alpha -\beta)=\frac{2}{5}+\frac{3\sqrt{3} }{10}=\frac{4+3\sqrt{3} }{10}$

Since $sin(\alpha)=\frac{4}{5}$

then $\alpha=arcsin(\frac{4}{5} )\approx 53.1^o$

4 0

3 years ago

What's the pattern rule for 8 12 24 60 168?

lilavasa [31]

He changes are 3 times larger each time.

8 changed by 4 to get to 12.

12 changed by 12 to get to 24.

24 changed by 36 to get to 60.

60 changed by 108 to get to 168.

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4 years ago