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klio [65]
2 years ago
11

Help please with this question!!!

Mathematics
1 answer:
Vesnalui [34]2 years ago
7 0
1 - AB is the Hypotenuse, BC is the opposite, AC is the adjacent (next to)

AB is the Hypotenuse because it is across from the right angle. BC is the opposite because its on the opposite side of the angle. AC is next to (adjacent) of the angle.

2 - the ratio is 30/1 = 5/b

This is because S O/H C A/H T O/A or Sohcahtoa.

S is the sine, which is Opposite over Hypotenuse. C is the cosine, which is Adjacent over Hypotenuse, and T is tangent which is Opposite over Adjacent. So look at what we labeled, we need to find the adjacent side, or b, and we have the opposite side, 5. So what has opposite and adjacent? Tangent.

So we do tan(30)/1 = 5/b, and cross multiply.

Now the last question. Hypotenuse is always C. But the other two values don’t matter where you put them on the left side since its multiplication. So its 7.7^2 * 6^2 = c^2 which is 59.29 * 36 = C^2.
2134.44 = c^2. Square root sides so it is 46.2 = c. The answer is 46.2
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Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).
ddd [48]

Answer:

f'(x) = 4x^2 - 3 for x \le -3

Step-by-step explanation:

See attachment for proper question

Given

f(x) = -\frac{1}{2}\sqrt{x + 3}

For

x \ge -3

Required

Determine the inverse function

f(x) = -\frac{1}{2}\sqrt{x + 3}

Replace f(x) with y

y = -\frac{1}{2}\sqrt{x + 3}

Swap the positions of x and y

x = -\frac{1}{2}\sqrt{y + 3}

Multiply both sides by -2

-2 * x =-2 *  -\frac{1}{2}\sqrt{y + 3}

-2x =\sqrt{y + 3}

Square both sides

(-2x)^2 =(\sqrt{y + 3})^2

4x^2 =y + 3

Make y the subject

y = 4x^2 - 3

The inverse has been solved. So, we need to replace y with f'(x)

f'(x) = 4x^2 - 3

Next, is to determine the interval

x \ge -3

Change inequality to \le

x \le -3

Hence, the inverse function is:

f'(x) = 4x^2 - 3 for x \le -3

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3 years ago
Where to put this on the number line? ​
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Answer:

Step-by-step explanation:

add 9 to both sides.
then divide by 1.5

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number your number line: 2,4,6,8,10

draw a circle at 8 and the line to the right since p>8

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