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34kurt
3 years ago
10

A right triangle has side lengths 4 units, 5units, and x units. It is unknown if the missing length is the longest or shortest s

ide. Rounded to the nearest tenth, what is the difference between the possible values of x?

Mathematics
2 answers:
svet-max [94.6K]3 years ago
8 0

Answer:

3rd Option is correct.

Step-by-step explanation:

Given: A right angles triangle.

          length of the sides of triangle are 4 units , 5 units and x units.

we have to find difference between possible value of x.

We use Pythagoras theorem of right angled triangle to fins the value of x.

Pythagoras theorem states that square of the hypotenuse is equal to the sum of the square of the legs of the triangle.  

If x is longest side that is hypotenuse.

then using Pythagoras theorem,  

x² = 4² + 5²

x² = 16 + 25

x² = 41

x = √41

x = 6.40 units

If x is shortest side ⇒ 5 is length of hypotenuse.

then using Pythagoras theorem,  

5² = 4² + x²

25 = 16 + x²

x² = 25 - 16

x² = 9

x = √9

x = 3 units

Difference of x = 6.40 - 3 = 3.40 unit

Therefore, 3rd Option is correct.

lapo4ka [179]3 years ago
7 0

Answer:

6.4

Step-by-step explanation:

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Write an algebraic
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x = 115 - p

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generally in algebra you will use the letter "x" to represent a value you do not know, so it wants an algebraic expression for "p subtracted from 115", this can be rewritten as 115 - p. So the unknown value  "x" is equal to 115 - p.

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What type of transformation transforms (a, b) to (-a, b)?
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A reflection over the y axis .

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Given  : Transforms (a,b) to (-a,b).

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Transformed  : (-a ,b).

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If 2/x+ 3/y =13 and 5/x- 4/y = -2, then x+y equals
creativ13 [48]

Answer:

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Step-by-step explanation:

Let:

u =  \frac{1}{x}  \\ v =  \frac{1}{y}

This allows us to manipulate the equations like we normally would.

2u + 3v = 13 \\ 5u - 4v =  - 2 \\  \\   + 8u + 12v = 52 \\  + 15u - 12v =  - 6 \\ 23u = 46 \\ u = 2 \\  \\ 2(2) + 3v = 13 \\ 4 + 3v = 13 \\ 3v = 9 \\ v = 3

Then, we return the values where they belong.

u =  \frac{1}{x}  \\ 2 =  \frac{1}{x}  \\ x =  \frac{1}{2}  \\  \\ v =  \frac{1}{y}  \\ 3 =  \frac{1}{y}  \\ y =  \frac{1}{3}

Finally, we add:

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