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abruzzese [7]
2 years ago
8

The median weight for a 5 foot tall male to enlist in the US Army is 114.5 lbs. This weight can vary by 17.5 lbs. Write and solv

e an absolute value inequality to represent the weight of a 5 foot male who would not meet the minimum or maximum weight requirement allowed to enlist the Army.
Mathematics
1 answer:
myrzilka [38]2 years ago
4 0

An absolute value inequality that represents the weight of a 5-foot male who would not meet the minimum or maximum weight requirement allowed to enlist in the Army is 97 lbs < x < 132 lbs.

<h3>What are inequalities?</h3>

Inequalities help us to compare two unequal expressions. Also, it helps us to compare the non-equal expressions so that an equation can be formed.

It is mostly denoted by the symbol <, >, ≤, and ≥.

The median weight for a 5 foot tall male to enlist in the US Army is 114.5 lbs. This weight can vary by 17.5 lbs. Therefore, the inequality can be written as,

(114.5 - 17.5) lbs < x < (114.5 + 17.5) lbs

97 lbs < x < 132 lbs

Hence, an absolute value inequality that represents the weight of a 5-foot male who would not meet the minimum or maximum weight requirement allowed to enlist in the Army is 97 lbs < x < 132 lbs.

Learn more about Inequality:

brainly.com/question/19491153

#SPJ1

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sergey [27]
The answer I got was 75%
5 0
3 years ago
What is the area of the shaded region in the figure below ? Leave answer in terms of pi and in simplest radical form
ser-zykov [4K]

Answer:

Step-by-step explanation:

That shaded area is called a segment. To find the area of a segment within a circle, you first have to find the area of the pizza-shaped portion (called the sector), then subtract from it the area of the triangle (the sector without the shaded area forms a triangle as you can see). This difference will be the area of the segment.

The formula for the area of a sector of a circle is:

A_s=\frac{\theta}{360}*\pi r^2 where our theta is the central angle of the circle (60 degrees) and r is the radius (the square root of 3).

Filling in:

A_s=\frac{60}{360}*\pi (\sqrt{3})^2 which simplifies a bit to

A_s=\frac{1}{6}*\pi(3) which simplifies a bit further to

A_s=\frac{1}{2}\pi which of course is the same as

A_s=\frac{\pi}{2}

Now for tricky part...the area of the triangle.

We see that the central angle is 60 degrees. We also know, by the definition of a radius of a circle, that 2 of the sides of the triangle (formed by 2 radii of the circle) measure √3. If we pull that triangle out and set it to the side to work on it with the central angle at the top, we have an equilateral triangle. This is because of the Isosceles Triangle Theorem that says that if 2 sides of a triangle are congruent then the angles opposite those sides are also congruent. If the vertex angle (the angle at the top) is 60, then by the Triangle Angle-Sum theorem,

180 - 60 = 120, AND since the 2 other angles in the triangle are congruent by the Isosceles Triangle Theorem, they have to split that 120 evenly in order to be congruent. 120 / 2 = 60. This is a 60-60-60 triangle.

If we take that extracted equilateral triangle and split it straight down the middle from the vertex angle, we get a right triangle with the vertex angle half of what it was. It was 60, now it's 30. The base angles are now 90 and 60. The hypotenuse of this right triangle is the same as the radius of the circle, and the base of this right triangle is \frac{\sqrt{3} }{2}. Remember that when we split that 60-60-60 triangle down the center we split the vertex angle in half but we also split the base in half.

Using Pythagorean's Theorem we can find the height of the triangle to fill in the area formula for a triangle which is

A=\frac{1}{2}bh. There are other triangle area formulas but this is the only one that gives us the correct notation of the area so it matches one of your choices.

Finding the height value using Pythagorean's Theorem:

(\sqrt{3})^2=h^2+(\frac{\sqrt{3} }{2})^2 which simplifies to

3=h^2+\frac{3}{4} and

3-\frac{3}{4}=h^2 and

\frac{12}{4} -\frac{3}{4} =h^2 and

\frac{9}{4} =h^2

Taking the square root of both the 9 and the 4 (which are both perfect squares, thankfully!), we get that the height is 3/2. Now we can finally fill in the area formula for the triangle!

A=\frac{1}{2}(\sqrt{3})(\frac{3}{2}) which simplifies to

A=\frac{3\sqrt{3} }{4}

Therefore, the area in terms of pi for that little segment is

A_{seg}=\frac{\pi}{2}-\frac{3\sqrt{3} }{4}, choice A.

8 0
3 years ago
Solve this equation for x. (x+4) (x-2)=0
-BARSIC- [3]
1. -4
2. 2
four plus negative four is zero and two minus two is zero
7 0
3 years ago
What is the LCM of 24 and 54?
Fiesta28 [93]

Answer: 216

Step-by-step explanation:

The LCM of 24 and 54 is 216. To find the least common multiple of 24 and 54, we need to find the multiples of 24 and 54 (multiples of 24 = 24, 48, 72, 96 . . . . 216; multiples of 54 = 54, 108, 162, 216) and choose the smallest multiple that is exactly divisible by 24 and 54, i.e., 216.

4 0
2 years ago
Read 2 more answers
This id hard please help me do this
svlad2 [7]
In this problem, we're going to use the 'PEMDAS' method.

First, we're doing the problem in the parenthesis.
44 - 4 = 40

Next in PEMDAS, we need to do the exponent.
6² = 6 x 6 = 36

Then we would do multiplication, but there is nothing to multiply so we move onto division.
40 / 2 = 20

Now we would add, but there is nothing to add so we move onto subtraction.
20 - 36 = -16

Our final answer would be -16.

Hope this helps!~
7 0
2 years ago
Read 2 more answers
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