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2 years ago

( a+b )^2 = ...........................

Mathematics

2 answers:

stepladder [879]2 years ago

5 0

Answer:

$a^{2} + 2ab +b^{2}$

svlad2 [7]2 years ago

5 0

answer: a^2+2ab+b^2

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Solve for the unknown quantity of x 15/2= 60/x

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Directions: Write the name and composition of the following composer/s below. Space provided for your answer.

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3 years ago

Solve for x. 5x=15+5

Travka [436]

Answer: x=4

Step-by-step explanation:

5x=15+5

15+5=20

5x=20

divide 5 on both sides

20/5=4

x=4

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Out of a box 200 apples 10 of them has gotten bad

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Step-by-step explanation:

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Ivan used coordinate geometry to prove that quadrilateral EFGH is a square.

Gelneren [198K]

Answer:

(A)Segment EF, segment FG, segment GH, and segment EH are congruent

Step-by-step explanation:

Step 1

Quadrilateral EFGH with points E(-2,3), F(1,6), G(4,3), H(1,0)

Step 2

Using the distance formula

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Given E(-2,3), F(1,6)

$|EF|=\sqrt{(6-3)^2+(1-(-2))^2}=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt{2}$

Given F(1,6), G(4,3)

$|FG|=\sqrt{(3-6)^2+(4-1)^2}=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt{2}$

Given G(4,3), H(1,0)

$|GH|=\sqrt{(0-3)^2+(1-4)^2}=\sqrt{(-3)^2+(-3)^2}=\sqrt{18}=3\sqrt{2}$

Given E (−2, 3), H (1, 0)

$|EH|=\sqrt{(0-3)^2+(1-(-2))^2}=\sqrt{(-3)^2+(3)^2}=\sqrt{18}=3\sqrt{2}$

Step 3

Segment EF ,E (−2, 3), F (1, 6)

Slope of |EF|= $\frac{6-3}{1+2} =\frac{3}{3}=1$

Segment GH, G (4, 3), H (1, 0)

Slope of |GH| $= \frac{0-3}{1-4} =\frac{-3}{-3}=1$

Step 4

Segment EH, E(−2, 3), H (1, 0)

Slope of |EH| $= \frac{0-3}{1+2} =\frac{-3}{3}=-1$

Segment FG, F (1, 6,) G (4, 3)

Slope of |EH| $=\frac{3-6}{4-1} =\frac{-3}{3}=-1$

Step 5

Segment EF and segment GH are perpendicular to segment FG.

The slope of segment EF and segment GH is 1. The slope of segment FG is −1.

Step 6

Segment EF, segment FG, segment GH, and segment EH are congruent.

The slope of segment FG and segment EH is −1. The slope of segment GH is 1.

Step 7

All sides are congruent, opposite sides are parallel, and adjacent sides are perpendicular. Quadrilateral EFGH is a square

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3 years ago

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Can someone plzzz help me with this math practice paper

weeeeeb [17]

Answer:

45 degrees for 8

135 degrees for 9

48 for 10

Yes a square is a rectangle

Next ones.

Sut = 21 becuase x=5

Step-by-step explanation:

Last one:

x^2 + 8 = 3x + 36

- 8 - 8

x^2 = 3x + 28

-3x -3x

x^2 - 3x = 28

(x · x) - 3x = 28.

This was were a little guess work was used,

I found that any number lower than 7 is less than 28 when pluged into x and any above is higher.

Hence x = 7

x^2 + 8 = 7^2 + 8

7 x 7 = 49. 49 + 8 = 57.

and

3x+36 = 7 x 3 + 36

7x3 = 21. 21 + 36 = 57.

Both lines are equal so x is indeed 7.

The RSTU rectangle

3x+6 = 5x-4

+4 +4

3x+10 = 5x

-3x -3x

10 = 2x

10/2 = 5

5 = x or x = 5

plug it in now

3 x 5 = 15. 15 + 6 = 21

and

5 x 5 = 25. 25 - 4 = 21

so x = 5

8-10

QRS = 45 degrees because bisects the square with a diagonal line from corner to corner

PTQ is a 135 degrees because it is wider than a 90 degrees angle and meets both upper corner from the middle of the square making it 135 degrees.

SQ = 48 because RT = 24 and RT is half the length of SQ meaning its length would be 48

SQ= 24 degrees because RT = 24 and if RT was to continue on the line it is on it will reach the length of SQ.

3 0

3 years ago