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2 years ago

15

A sandwich shop offers three types of bread, four types of meat, and two types of cheese. How many different sandwiches can be m

ade, using one type of
bread, one type of meat, and one type of cheese?
A.9
B. 14
C.21
D. 24

1 answer:

fomenos2 years ago

6 0

my Answer is (A) hbu I hope it did help you

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Molodets [167]

Answer:

198.00

Step-by-step explanation:

I am not sure but I hope this helps some.

8 0

2 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

There are three clubs at Redwood High School: Debate, Student council, and Key club. There are 1000 students in the school. And

hram777 [196]

Answer:

The number of students in the key club is 490

Step-by-step explanation:

The given parameters are;

The number of student in the school = 1000

The total number of student in the debate club = 310

The total number of student in the student council = 650

The total number of student who are in debate and student council = 170

The total number of student who are in both debate and the key club = 150

The total number of student who are in both student council and the key club = 180

The number of students who are in all three clubs = 50

Therefore, we have;

Let A represent the number of students in the debate club

Let B represent the number of students in the student council

Let C represent the number of students in the key club

We have;

n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) - n(C∩A) + n(A∩B∩C)

Where;

n(A∪B∪C) = 1000

n(A) = 310

n(B) = 650

n(A∩B) = 170

n(B∩C) = 180

n(C∩A) = 150

n(A∩B∩C) = 50

Therefore;

n(C) = n(A∪B∪C) - (n(A) + n(B) - n(A∩B) - n(B∩C) - n(C∩A) + n(A∩B∩C))

Substituting the values gives;

n(C) = 1000 - (310 + 650 - 170 - 180 - 150 + 50) = 490

The number of students in the key club, n(C) = 490.

5 0

3 years ago

Classy the polynomial 5x^2+4x-2 by degree

bogdanovich [222]

5x^2+4x-2 This is a second degree polynomial since the largest exponent is x^2

5 0

3 years ago

Read 2 more answers

Wesleys mother is 57 years old. her age is three years more than twice edwins age. what is wesleys age?

natita [175]

Answer: Edwin is likely a typo for Wesley.

Step by step:

Wmom=57

Wmom=3+2e

57=3+2e

54=2e

27=e

Edwin is 27 years old.

Aside from a typo, one can speculate on Wesley's age. He's likely between 20 and 39 years old based on his mother's age. If you want to delve into societal norms, Wesley is likely to be a name from a white middle to upper class parentage, so it's less likely he was had when his mother was a teen, or even in her early twenties because college, so move that age zone to 20 to 34. Then consider that having children later than 35 is consideration for a "risky" pregnancy and most women are urged to aim for before 33, but most upper middle class white women will be pressured to baby before 30. So, adjust Wesley's age zone to 27 to 34. And then...idk. But in all honesty I think Edwin is meant to be Wesley, because it's a math problem not a sociology one.

7 0

3 years ago