Help me find y please

If the length of a rectangle is increased by 120 %, and its width is decreased by 20%,

Dima020 [189]

Let length of rectangle=x and width be y
.`. area=x*y
now new length=x+12x/10=22x/10
width=y-y/5=4y/5
now .....
new area=new length-new width=22x/10*4y/5
new area=88xy/50

7 0

3 years ago

Read 2 more answers

Use the Polynomial Identity below to help you create a list of 10 Pythagorean Triples:

Stolb23 [73]

Answer:

(3,4,5)

(6,8,10)

(5,12,13)

(8,15,17)

(12,16,20)

(7,24,25)

(10,24,26)

(20,21,29)

(16,30,34)

(9,40,41)

Just choose 2 numbers from {1,2,3,4,5,6,7,8,...} and make sure the one you input for x is larger.

Post the three in the comments and I will check them for you.

Step-by-step explanation:

We need to choose 2 positive integers for x and y where x>y.

Positive integers are {1,2,3,4,5,6,7,.....}.

I'm going to start with (x,y)=(2,1).

x=2 and y=1.

$(2^2+1^2)^2=(2^2-1^2)^2+(2\cdot2\cdot1)^2$

$(4+1)^2=(4-1)^2+(4)^2$

$(5)^2=(3)^2+(4)^2$

So one Pythagorean Triple is (3,4,5).

I'm going to choose (x,y)=(3,1).

x=3 and y=1.

$(3^2+1^2)^2=(3^2-1^2)^2+(2\cdot3\cdot1)^2$

$(9+1)^2=(9-1)^2+(6)^2$

$(10)^2=(8)^2+(6)^2$

So another Pythagorean Triple is (6,8,10).

I'm going to choose (x,y)=(3,2).

x=3 and y=2.

$(3^2+2^2)^2=(3^2-2^2)^2+(2\cdot3\cdot2)^2$

$(9+4)^2=(9-4)^2+(12)^2$

$(13)^2=(5)^2+(12)^2$

So another is (5,12,13).

I'm going to choose (x,y)=(4,1).

$(4^2+1^2)^2=(4^2-1^2)^2+(2\cdot4\cdot1)^2$

$(16+1)^2=(16-1)^2+(8)^2$

$(17)^2=(15)^2+(8)^2$

Another is (8,15,17).

I'm going to choose (x,y)=(4,2).

$(4^2+2^2)^2=(4^2-2^2)^2+(2\cdot4\cdot2)^2$

$(16+4)^2=(16-4)^2+(16)^2$

$(20)^2=(12)^2+(16)^2$

We have another which is (12,16,20).

I'm going to choose (x,y)=(4,3).

$(4^2+3^2)^2=(4^2-3^2)^2+(2\cdot4\cdot3)^2$

$(16+9)^2=(16-9)^2+(24)^2$

$(25)^2=(7)^2+(24)^2$

We have another is (7,24,25).

You are just choosing numbers from the positive integer set {1,2,3,4,... } and making sure the number you plug in for x is higher than the number for y.

I will do one more.

Let's choose (x,y)=(5,1).

$(5^2+1^2)^2=(5^2-1^2)^2+(2\cdot5\cdot1)^2$

$(25+1)^2=(25-1)^2+(10)^2$

$(26)^2=(24)^2+(10)^2$

So (10,24,26) is another.

Let (x,y)=(5,2).

$(5^2+2^2)^2=(5^2-2^2)^2+(2\cdot5\cdot2)^2$

$(25+4)^2=(25-4)^2+(20)^2$

$(29)^2=(21)^2+(20)^2$

So another Pythagorean Triple is (20,21,29).

Choose (x,y)=(5,3).

$(5^2+3^2)^2=(5^2-3^2)^2+(2\cdot5\cdot3)^2$

$(25+9)^2=(25-9)^2+(30)^2$

$(34)^2=(16)^2+(30)^2$

Another Pythagorean Triple is (16,30,34).

Let (x,y)=(5,4)

$(5^2+4^2)^2=(5^2-4^2)^2+(2\cdot5\cdot4)^2$

$(25+16)^2=(25-16)^2+(40)^2$

$(41)^2=(9)^2+(40)^2$

Another is (9,40,41).

5 0

3 years ago

A present is shaped like a

matrenka [14]

Answer: Yes

Step-by-step explanation:

8 0

3 years ago

HELP ME PLZ and this asks to show your work so plz do

skelet666 [1.2K]

The surface area of a three dimensional object is the sum of the areas of all its faces. Here we have a net of triangular prism, which if we were to fold would form a three dimensional shape.

We need to find the area of each face. Let's begin with the rectangle in the center which can be found by calculating its length times width:

26 * 10 = 260 in^2

Next, let's find the area of the other two rectangles. Although it does not specify that these rectangles are congruent (meaning the same), we know that they are because if they were different sizes, the prism would not fit together when folded. We can find the area of one rectangle and multiply by two:

26 * 13 = 338

338 * 2 = 676 in^2

Lastly, we have two triangles which are congruent for the same reason as the rectangles. The area of a triangle can be found by calculating one-half the base time the height:

0.5 * 10 * 12 = 60

Times two since there are two triangles:

60 * 2 = 120 in^2

Add up the areas:

260 + 676 + 120 = 1056 in^2

Multiply the surface area by 6 because there are 6 packages:

1056 * 6 = 6336 in^2

The answer is C.

8 0

3 years ago

Find the lowest common multiple ( L.C.M. ) of the (a² - 4), (a³ - 8), ( a + 2 ) ²

Murljashka [212]

Answer: according to my work i got

LCM - Lowest Common Factor

HCF - Highest common factor

Step-by-step explanation:

please mark brainlest!!

yw! =^.^=

6 0

2 years ago

Help me find y please​

Help me find y please