How would I factor #4?

Which of the following inequalities matches the graph?

Neko [114]

Answer:

c

Step-by-step explanation:

7 0

3 years ago

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PLEASE HELP!!! ITS URGENT AND ILL NEED IT SOON, THANK YOU FOR ANY HELP GIVEN IN ADVANCED COMES WITH DIAGRAM

andrew11 [14]

Answer:

Try 458.96 (round if needed)

Step-by-step explanation:

First you find the volume of the cone. (pi*radius^2*h/3)

Then you find the volume of a cylinder. (pi*radius^2*h) add these two up.

You would have to subtract 12.5 and 8.5 to get the height of the cone.

Also you would have to subtract 6 inches off of your total.

(Sorry If this explanation sucked.)

6 0

3 years ago

Suppose a casino invents a new game that you must pay $250 to play. The game works like this: The casino drawsrandom numbers bet

Ira Lisetskai [31]

Answer:

clc%clears screen

clear all%clears history

close all%closes all files

p=250;

M=[];

for i=1:100000

re=0;

S=0;

while(S<=1)

S=S+rand;

re=re+100;

end

M(i)=re;

end

disp('Expected received money is');

mean(M)

disp('Since expcted is greater than what we pay. So, we will play')

Step-by-step explanation:

6 0

3 years ago

What is the probability that 2 randomly selected months have 31 days?

aniked [119]

Answer:

31.8%

Step-by-step explanation:There are 7 months that have 31 days so,

The odds the first selected month has 31 days is 7/12

The next month there are only 6 months left with 31 days of only 11 months to choose from

so the odds for the second are 6/11

To calculate the odds they BOTH have 31 months you multiply the two odds:

(7/12) x (6/11) = 42/132 = .318 so 31.8%

6 0

3 years ago

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Which series of transformations will not map figure H onto itself

7nadin3 [17]

Answer:

D

Step-by-step explanation:

Given a square with vertices at points (2,1), (1,2), (2,3) and (3,2).

Consider option A.

1st transformation $(x+0,y-2)$ will map vertices of the square into points

$(2,1)\rightarrow (2,-1);$
$(1,2)\rightarrow (1,0);$
$(2,3)\rightarrow (2,1);$
$(3,2)\rightarrow (3,0).$

2nd transformation = reflection over y = 1 has the rule (x,2-y). So,

$(2,-1)\rightarrow (2,3);$
$(1,0)\rightarrow (1,2);$
$(2,1)\rightarrow (2,1);$
$(3,0)\rightarrow (3,2)$

These points are exactly the vertices of the initial square.

Consider option B.

1st transformation $(x+2,y-0)$ will map vertices of the square into points

$(2,1)\rightarrow (4,1);$
$(1,2)\rightarrow (3,2);$
$(2,3)\rightarrow (4,3);$
$(3,2)\rightarrow (5,2).$

2nd transformation = reflection over x = 3 has the rule (6-x,y). So,

$(4,1)\rightarrow (2,1);$
$(3,2)\rightarrow (3,2);$
$(4,3)\rightarrow (2,3);$
$(5,2)\rightarrow (1,2)$

These points are exactly the vertices of the initial square.

Consider option C.

1st transformation $(x+3,y+3)$ will map vertices of the square into points

$(2,1)\rightarrow (5,4);$
$(1,2)\rightarrow (4,5);$
$(2,3)\rightarrow (5,6);$
$(3,2)\rightarrow (6,5).$

2nd transformation = reflection over y = -x + 7 will map vertices into points

$(5,4)\rightarrow (3,2);$
$(4,5)\rightarrow (2,3);$
$(5,6)\rightarrow (1,2);$
$(6,5)\rightarrow (2,1)$

These points are exactly the vertices of the initial square.

Consider option D.

1st transformation $(x-3,y-3)$ will map vertices of the square into points

$(2,1)\rightarrow (-1,-2);$
$(1,2)\rightarrow (-2,-1);$
$(2,3)\rightarrow (-1,0);$
$(3,2)\rightarrow (0,-1).$

2nd transformation = reflection over y = -x + 2 will map vertices into points

$(-1,-2)\rightarrow (4,3);$
$(-2,-1)\rightarrow (3,4);$
$(-1,0)\rightarrow (2,3);$
$(0,-1)\rightarrow (3,2)$

These points are not the vertices of the initial square.

5 0

3 years ago