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andre [41]
3 years ago
14

I don't understand how to do find x and y, please help? thank yous.

Mathematics
1 answer:
pav-90 [236]3 years ago
7 0

#8) There are THREE right-angle (90°) TRIANGLES.

The "big" triangle has legs "x" and "no designation - <em>so lets call it Z" </em>and hypotenuse "4+5"<em> = 9</em>

The "left" triangle has legs "4" and "y" and hypotenuse "x"

The "right" triangle has legs "5" and "y" and hypotenuse "no designation - <em>which we named Z</em>"

Now, you can create a system of equations - you have three unknown variables and can create an equation for each triangle using the Pythagorean Theorem.

Big: x² + Z² = 9² → x² + Z² = 81

Left: 4² + y² = x² → 16 + y² = x² → 16 = x² - y² → x² - y² = 16

Right: 5² + y² = Z² → 25 + y² = Z²

Next, let's eliminate Z by using substitution....use the Right equation to plug Z² into the Big equation to create a NEW Big equation.

NEW Big: x² + (25 + y² ) = 81 → x² + y² = 81 - 25 → x² + y² = 56

Now, we have two unknown variables (x and y) and two equations (NEW Big and Left). Solve the system using any method. <em>I am going to solve using the elimination method)</em>

NEW Big: x² + y² = 56

+ Left: <u>x² - y² = 16</u>

2x² = 72 <em>added the NEW Big and Left equations</em>

x² = 36 <em>divided both sides by 2</em>

x = 6 <em>took square root of both sides (NOTE: -6 is disregarded because we are solving for a length, which cannot be negative) </em>

The last step is to plug the "x" value (<em>that we just solved for)</em> into one of the equations to solve for "y". <em>I am choosing to use the Left equation.</em>

<em> </em>x² - y² = 16

36 - y² = 16<em> </em>

- y² = 16 - 36

- y² = -20

y² = 20

y = √20

y = 2√5

Answer: x = 6, y = 2√5

Try #9 on your own and then check your answers below:

Big: Z² + x² = 12², Left: 3² + y² = Z², Right: y² + 9² = x²

Answer: x = 6√3, y = 3√3

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astraxan [27]

The right answer is Option A.

Step-by-step explanation:

Let,

x be the postcards

y be the large envelops

According to given statement;

14x+5y=12   Eqn 1

10x+15y=24.80   Eqn 2

Multiplying Eqn 1 by 3;

3(14x+5y=12)\\42x+15y=36\ \ \ Eqn \ 3

Subtracting Eqn 2 from Eqn 3;

(42x+15y)-(10x+15y)=36-24.80\\42x+15y-10x-15y=11.20\\32x=11.20\\

Dividing both sides by 32

\frac{32x}{32}=\frac{11.20}{32}\\x=0.35

Putting x=0.35 in Eqn 1;

14(0.35)+5y=12\\4.90+5y=12\\5y=12-4.90\\5y=7.10

Dividing both sides by 5

\frac{5y}{5}=\frac{7.10}{5}\\y=1.42

Therefore, one large envelope costs $1.42

The right answer is Option A.

Keywords: linear equations, subtraction

Learn more about linear equations at:

  • brainly.com/question/9738996
  • brainly.com/question/10015690

#LearnwithBrainly

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