MATHEMATICS HIGH SCHOOL

The radius of the base of a cylinder is 10 centimeters, and its height is 20 centimeters. A cone is used to fill the cylinder with water. The radius of the cone's base is 5 centimeters, and its height is 10 centimeters.The number of times one needs to use the completely filled cone to completely fill the cylinder with water is.....

Answers

Answer 1

Answer:

Answer:

24 times.

Step-by-step explanation:

Since, the volume of a cone,

$V=(1)/(3)\pi r^2 h$

While, the volume of a cylinder,

$V=\pi r^2 h$

Where, r = radius,

h = height,

Thus, the volume of the cone having radius 5 cm and height 10 cm,

$V_1=(1)/(3)\pi (5)^2(10)$

And, the volume of the cylinder having radius 10 cm, and 20 cm,

$V_2=\pi (10)^2 (20)$

Hence, the number of times we need to use cone to completely fill the cylinder = $(V_2)/(V_1)$

$=(\pi (10)^2 (20))/((1)/(3)\pi (5)^2(10))$

$=(2000)/(250/3)$

$=(6000)/(250)$

$=24$

Answer 2

Answer: Im like 90% sure that your answer should be 2.

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Explain why the sine of an acute angle of a right triangle is equal to the cosine of its complement

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to see this relationship, all you need to know is some basic trig.

consider triangle ABC. the sin(β)=b/c ; cos(α)=b/c, therefore sin(β)=cos(α). also, since this is a right triangle, β and α are complements

Solve for x₁ and x₂
6x₁+3x₂=60
6x₁-2x₂=80

Answers

$\left \{ {\big{6x_1+3x_2=60\ \ \ \ \ \ \ \ \ \ \ } \atop \big{6x_1-2x_2=80\ /\cdot(-1)}} \right. \n\n \left \{ {\big{6x_1+3x_2=60\ \ \ \ } \atop \big{-6x_1+2x_2=-80}} \right. \n-----------\n3x_2+2x_2=60-80\n5x_2=-20\ /:5\n\nx_2=-4\n\n6x_1+3x_2=60\ /:3\n2x_1+x_2=20\ \ \ \Rightarrow\ \ \ 2x_1+(-4)=20\ \ \ \Rightarrow\ \ \ 2x_1=24\ /:2\n\nx_1=12\n\n \left \{ {\big{x+1=12} \atop\big {x_2=-4}} \right.$

first let's change the "x1" to "A" Nd the "x2" to "B" __ so now the problem reads "6A + 3B = 60" And "6A - 2B= 80" we want to isolate variable. and the easiest one is the "B" by getting rid of the "A". set up a subtraction problem where you take one equation away from the other. 6a-6a=0. 3b - (-2b) = 5b and 60-80=-20. so your new equation reads 5B= -20. you wanna get B on one side so yu divide 5 on both sides giving you b= -4 or "x2"=-4. from there you can plug -4 into any one of the original equations. ex: 6A -2(-4)=80 then 6A + 8=80 then you subtract 8 from both sides because you're trying to isolate the "A" variable. which leaves you with 6A = 72. and now you wanna get"A" on one side so yu divide 6 from both sides and get A=12 or "x1"=12. do a quick check and plug in these numbers into any one of the equation.

How many ways are there to select a committee to develop a discrete mathematics course at a school if the committee is to consist of 3 faculty members from the mathematics department and 4 from the computer science department, if there are 9 faculty members of the math department and 11 of the CS department?

Answers

Answer:

There are 27,720 ways to select the committee

Step-by-step explanation:

First, it is necessary to know how many ways are there to select 3 members, if there are 9 members of the mathematics department. This can be found using the following equation:

$nCk=(n!)/(k!(n-k)!)$

Where nCk gives as the number of ways in which we can select k elements from a group of n elements. So, replacing n by 9 and k by 3 members, we get:

$9C3=(9!)/(3!(9-3)!)=84$

So, there are 84 ways to select 3 members from 9 members of the mathematics department.

At the same way, we can calculate that there are 330 ways to select 4 members from the 11 that belong to the Computer science department as:

$11C4=(11!)/(4!(11-4)!)=330$

Finally the total number of ways in which we can form a committee with 3 faculty members from mathematics and 4 from the computer science department is calculated as:

9C3 * 11C4 = 84 * 330 = 27,720

Perform the indicated operation.

4/11 · 10/8

4/11
5/11
6/11

Answers

4/11 * 10/8

Multiplying fractions
1) multiply the numerators: 4 * 10 = 40
2) multiply the denominators: 11 * 8 = 88
3) simplify the fraction
40/88 ⇒ 10/22 ⇒ 5/11

40 ÷ 4 = 10
88 ÷ 4 = 22

10 ÷ 2 = 5
22 ÷ 2 = 11

4/11 * 10/8 = 5/11

There are 125 people and three door prizes. How many ways can three door prizes of $50 each be distributed? How many ways can door prizes of $5,000, $500 and $50 be distributed?

Answers

We have been given that there are 125 people and three door prizes.

In the first part we need to figure out how many ways can three door prizes of $50 each be distributed?

Since there are total 125 people and there are three identical door prices, therefore, we need to use combinations for this part.

Hence, the required number of ways are:

$_(3)^(125)\textrm{C}=(125!)/(122!3!)=(125*124*123)/(1*2*3)=317750$

In the next part, we need to figure out how many ways can door prizes of $5,000, $500 and $50 be distributed?

Since we have total 125 people and there are three prices of different values, therefore, the required number of ways can be figured out by using permutations.

$_(3)^(125)\textrm{P}=(125!)/(122!)=125*124*123=1906500$

Pls I need help !!!!!!!

Answers

Answer:

-3 -4

Step-by-step explanation:

Bc it's why not ...... ummmmmmm

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Cammile wants to make fruit drinks.The direction to make one drink include mixing 4 cup of yogurt and 1 cup of ice with the amount of each fruit shown8 5 cup of banana slice 8 2 cup of blueberries 8 Part A Camille wants to make six drinks for her friends.How many total cups of blueberries and banana slices will she use to make the 6 drinks?Show your work A:7 B:12 C:30 D:42 8. 8. 8. 8 Part B: Next Camille will add the yogurt and ice.how many yogurt and ice will she use to make 6 drinks?Show your work or explain your answer. Enter your answer and work or explanation in the space provided.

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