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ioda
1 year ago
8

Find a right triangle that has legs with your irrational Lengths and a hypotenuse with a rational Length

Mathematics
1 answer:
cluponka [151]1 year ago
4 0

gg easy


the legs of a triangle are related as a^2+b^2=c^2 wehre a, b, and c are the legs and the hypotonuse of a triangle respectively


let's find some stuff

one easy way is to find a rational value for c find values of a and b

lets pick c=5

a^2+b^2=5^2

a^2+b^2=25

now split up 25 into 2 things that don't have rational square roots

if we say a^2=19 and b^2=6 then we get

a=\sqrt{19} and b=\sqrt{6} (which are both irrational)

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610 people go on a coach trip each coach can hold 55 people how many coaches are needed
diamong [38]

Answer:

12

Step-by-step explanation:

610 divided by 55 = 11.09

but since u cant have 11.09, u need 12

4 0
2 years ago
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Question:twice as much as the sum of eighteen and eleven<br><br> in numbers and don't solve
Dmitrij [34]
2(18+11)
i think this is the expression you are looking for
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2 years ago
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19. Bella is putting down patches of sod
Fofino [41]

Answer:

The dimensions of the two different rectangular regions are;

1st Arrangement:

W = 4 yards and L = 5 yards or W = 5 yards and L = 4 yards

2nd Arrangement:

W = 2 yards and L = 10 yards or W = 10 yards and L = 2 yards

The perimeter of the two different rectangular regions are;

1st Arrangement:

P₁ = 18 yards

2nd Arrangement:

P₂ = 24 yards

Step-by-step explanation:

Bella is putting down patches of sod to start a new lawn.

She has 20 square yards of sod.

We are asked to provide the dimensions of two different rectangular regions that she can cover with the sod.

Recall that a rectangle has an area given by

Area = W*L

Where W is the width of the rectangle and and L is the length of the rectangle.

Since Bella has 20 square yards of sod,

20 = W*L

There are more than two such possible rectangular arrangements.

Out of them, two different possible arrangements are;

1st Arrangement:

20 = (4)*(5) = (5)*(4)

Width is 4 yards and length is 5 yards or width is 5 yards and length is 4 yards

2nd Arrangement:

20 = (2)*(10) = (10)*(2)

Width is 2 yards and length is 10 yards or width is 10 yards and length is 2 yards

Therefore, the dimensions of two  different rectangular regions are;

1st Arrangement:

W = 4 yards and L = 5 yards or W = 5 yards and L = 4 yards

2nd Arrangement:

W = 2 yards and L = 10 yards or W = 10 yards and L = 2 yards

What is the perimeter of each region?

The perimeter of a rectangular shape is given by

P = 2(W + L)

Where W is the width of the rectangle and and L is the length of the rectangle.

The perimeter of the 1st arrangement is

P₁ = 2(4 + 5)

P₁ = 2(9)

P₁ = 18 yards

The perimeter of the 2nd arrangement is

P₂ = 2(2 + 10)

P₂ = 2(12)

P₂ = 24 yards

So the perimeter of the 1st arrangement is 18 yards and the perimeter of the 2nd arrangement is 24 yards.

Note:

Another possible arrangement is,

20 = (1)*(20) = (20)*(1)

Width is 1 yard and length is 20 yards or width is 20 yards and length is 1 yard.

3 0
1 year ago
A sphere has a radius of 16 in. Which statements about the sphere are true? Check all that apply.
Wewaii [24]

Answer:

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  • The radius' lenght is one half the length of the diameter. CORRECT: , by definition, the diameter is the maximum distance between two opposite points on the surface of the sphere, and its lenght equals twice the radius because the radius is the distance between the center of the sphere and any point of the frontier of the sphere (then, one colud imagine that the distance between two opposite points in a sphere can be united by two radius, or a diameter).
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6 0
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vredina [299]
First, list the numbers from smallest to greatest:
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