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Rasek [7]
3 years ago
10

Three pieces of wood measure 20 cm, 41 cm, 44 cm. If the same amount is cut off each piece, the remaining length can be formed i

nto a right triangle. What is the length that is cut off?
Mathematics
1 answer:
SIZIF [17.4K]3 years ago
7 0

The length that can be cut off from the three given pieces of wood equally to form a right triangle is 5 cm

Given parameters:

dimension of the three pieces of wood = 20 cm, 41 cm and 44 cm

To find:

  • the length that is cut off to form a right triangle

let the length that is cut off from each of the wood = y

From Pythagoras theorem, we will have the following equation.

(44-y)^2 = (20-y)^2 + (41-y)^2\\\\expand \ the \ equation \ as \ follows;\\\\1936 - 88y + y^2 = 400 - 40y + y^2  + 1681 - 82y + y^2\\\\simplify \ by \ collecting \ similar \ terms \ together\\\\(1936 - 400- 1681) + (-88y + 40y+ 82y) + (y^2 - 2y^2) = 0\\\\-145 +34y - y^2 = 0\\\\multiply \ through \ by \ (-1)\\\\y^2-34y + 145 = 0\\\\factorize \ the \ above \ quadratic\  equation\\\\y^2 -5y - 29y + 145 = 0\\\\y(y - 5) -29(y - 5)=0 \\\\(y -5)(y-29) = 0\\\\y = 5 \ \ \ or \ \ \ \ y = 29

Since the least measurement of one of the pieces of the wood is 20 cm, we cannot cut off 29 cm.

Thus, the highest amount we can cut off equally is 5 cm

learn more here: brainly.com/question/15808950

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Three-fourths of a number increased by one-half is equal to fourteen. Find the number.
ivanzaharov [21]

Three-fourths of a number increased by one-half

= 3/4 x + 1/2

So three-fourths of a number increased by one-half is equal to fourteen:

3/4 x + 1/2 = 14

Multiply both sides by 4

3x + 2 = 56

Subtract 2 from both sides

3x = 54

Divide both sides by 3

x = 18

Answer: The number is 18

8 0
4 years ago
Read 2 more answers
Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th
grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

S = \{1,1,2,2,2,3\}

And the probability of each is:

P(1) = \frac{n(1)}{n(s)}

P(1) = \frac{2}{6}

P(1) = \frac{1}{3}

P(2) = \frac{n(2)}{n(s)}

P(2) = \frac{3}{6}

P(2) = \frac{1}{2}

P(3) = \frac{n(3)}{n(s)}

P(3) = \frac{1}{6}

P(Odd Number) is then calculated as:

P(Odd\ Number) =  P(1) + P(3)

P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}

Take LCM

P(Odd\ Number) = \frac{2+1}{6}

P(Odd\ Number) = \frac{3}{6}

P(Odd\ Number) =  \frac{1}{2}

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

S = \{1,1,2,2,2,3,3\}

The probability of each is:

P(1) = \frac{n(1)}{n(s)}

P(1) = \frac{2}{7}

P(2) = \frac{n(2)}{n(s)}

P(2) = \frac{3}{7}

P(3) = \frac{n(3)}{n(s)}

P(3) = \frac{1}{7}

P(Odd\ Number) = P(1) + P(3)

P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}

Take LCM

P(Odd\ Number) = \frac{2+1}{7}

P(Odd\ Number) = \frac{3}{7}

Comparing P(Odd Number) before and after

P(Odd\ Number) =  \frac{1}{2} --- Before

P(Odd\ Number) = \frac{3}{7} --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

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3 years ago
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meriva

Using the Quadratic formula

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Y’all I need a lot of help please help me and please show work
Nadya [2.5K]

Answer:

90

Step-by-step explanation:

no of total groups = 2+6+4=12

so in total there are 12 groups

each have equal number of members

so the answer is the number that is not a factor of 12

i hope this help you

this is quite high order thinking question

so don't know about the steps

if you understood then you may write the steps yourself

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A or B ....it’s your choice
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