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Leokris [45]
3 years ago
8

what could be a side length of a right triangle A.) 30 in, 45 in, 50 in B.) 30 in, 40 in,50 in C.) 30 in, 40 in, 60 in D.) 25 in

, 40 in, 50 in​
Mathematics
1 answer:
lions [1.4K]3 years ago
4 0

The pythagorean theorem holds for every right triangle: given the legs a, b and the hypothenuse c, the triangle is right if and only if

a^2+b^2=c^2

So, you have to check:

30^2+45^2=2925\neq 2500 = 50^2

So the first triangle can't be a right triangle.

30^2+40^2=2500= 50^2

So the second triangle is a right triangle.

The third triangle can't be right, because it has the same legs but a different hypothenuse

Finally, we have

25^2+40^2=2225= 50^2

So the last triangle can't be a right triangle.

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Points H, J, G, and K

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left bottom is 3

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G and K are on the line but that still makes them part of a Quadratic Just they get to be part of 2 of them like...

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3 years ago
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Use one transformation to solve n-4.6=2.98
Serggg [28]
What are the options of tranformation ??/
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Convert the decimal to a percent: 3.5​
Varvara68 [4.7K]
(3.5 × 100) × 1100 = 350100.
Write in percentage notation: 350%
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3 years ago
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Foe the AP given by a1,a2,......,an,...with non zero common difference, the equation satisfied are
user100 [1]
The given sequence is
a₁, a₂, ..., a_{n}

Because the given sequence is an arithmetic progression (AP), the equation satisfied is
a_{n}=a_{1}+(n-1)d
where
d =  the common difference.

The common difference may be determined as
d = a₂ - a₁

The common difference is the difference between successive terms, therefore
d = a₃ - a₂ = a₄ - a₃, and so on..

The sum of the first n terms is
S_{n}= \frac{n}{2} (a_{1}+a_{n})

Example:
For the arithmetic sequence
1,3,5, ..., 
the common difference is d= 3 - 1 = 2.
The n-th term is
a_{n}=1+2(n-1)

For example, the 10-term is
a₁₀ = 1 + (10-1)*2 = 19
Th sum of th first 10 terms is
S₁₀ = (10/2)*(1 + 19) = 100



8 0
3 years ago
50 POINTS!!! In rectangle ABCD, AB = 6 cm, BC = 8 cm, and DE = DF. The area of triangle DEF is one-fourth the area of rectangle
aalyn [17]

Answer:

EF=4\sqrt{3}

Step-by-step explanation:

In rectangle ABCD, AB = 6, BC = 8, and DE = DF.

ΔDEF is one-fourth the area of rectangle ABCD.

We want to determine the length of EF.

First, we can find the area of the rectangle. Since the length AB and width BC measures 6 by 8, the area of the rectangle is:

A_{\text{rect}}=8(6)=48\text{ cm}^2

The area of the triangle is 1/4 of this. Therefore:

\displaystyle A_{\text{tri}}=\frac{1}{4}(48)=12\text{ cm}^2

The area of a triangle is half of its base times its height. The base and height of the triangle is DE and DF. Therefore:

\displaystyle 12=\frac{1}{2}(DE)(DF)

Since DE = DF:

24=DF^2

Thus:

DF=\sqrt{24}=\sqrt{4\cdot 6}=2\sqrt{6}=DE

Since ABCD is a rectangle, ∠D is a right angle. Then by the Pythagorean Theorem:

(DE)^2+(DF)^2=(EF)^2

Therefore:

(2\sqrt6)^2+(2\sqrt6)^2=EF^2

Square:

24+24=EF^2

Add:

EF^2=48

And finally, we can take the square root of both sides:

EF=\sqrt{48}=\sqrt{16\cdot 3}=4\sqrt{3}

6 0
3 years ago
Read 2 more answers
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