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Alika [10]
3 years ago
12

Which of the following sets could be the sides of a right triangle?

Mathematics
1 answer:
coldgirl [10]3 years ago
7 0

Answer: Choice B) {3, 5, sqrt(34)}

=====================================

Explanation:

We can only have a right triangle if and only if a^2+b^2 = c^2 is a true equation. The 'c' is the longest side, aka hypotenuse. The legs 'a' and 'b' can be in any order you want.

-----------

For choice A,

a = 2

b = 3

c = sqrt(10)

So,

a^2+b^2 = 2^2+3^2 = 4+9 = 13

but

c^2 = (sqrt(10))^2 = 10

which is not equal to 13 from above. Cross choice A off the list.

-----------

Checking choice B

a = 3

b = 5

c = sqrt(34)

Square each equation

a^2 = 3^2 = 9

b^2 = 5^2 = 25

c^2 = (sqrt(34))^2 = 34

We can see that

a^2+b^2 = 9+25 = 34

which is exactly equal to c^2 above. This confirms the answer.

-----------

Let's check choice C

a = 5, b = 8, c = 12

a^2 = 25, b^2 = 64, c^2 = 144

So,

a^2+b^2 = c^2

25+64 = 144

89 = 144

which is a false equation allowing us to cross choice C off the list.

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Find the area. The figure is not drawn to scale
poizon [28]

Answer:

96 square inches.

Step-by-step explanation:

From given picture we see that

BC=8 in

AD=10 in

Given that

AD=AB

Then AB=10 in

Apply Pythagorean theorem in triangle ABC to find AC

(hypotenuse)^2=(base)^2+(perpendicular)^2

(AB)^2=(BC)^2+(AC)^2

(10)^2=(8)^2+(AC)^2

100=64+(AC)^2

100-64=(AC)^2

36=(AC)^2

take square root

6=AC

Then area of triangle ABC =\frac{1}{2}\left(base\right)\left(altitude\right)

=\frac{1}{2}\left(8\right)\left(6\right)=24

There are total 4 congruent triangles.

Then total area of the given figure = 4(24)=96 square inches.

7 0
3 years ago
Find the value of the following expression: (3^8 ⋅ 2^-5 ⋅ 90)−2 ⋅ ⋅ 3^28 (5 points) Write your answer in simplified form. Show a
Darya [45]
18452.8125 use PEMDAS
3 0
3 years ago
Is -3 greater or less than -5
OleMash [197]

Answer:

yes because its closer to not being negitibve

Step-by-step explanation:

3 0
3 years ago
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Find the fifth term of the binomial expansion.<br> (x2+y5) ^8
OlgaM077 [116]

Answer:

The fifth term of the binomial expansion

T_{5} = 70 x^{8} y^{20}

Step-by-step explanation:

<u><em>Step:1</em></u>

Given that the binomial expansion

     ( x² + y⁵ )⁸

we know that

T_{r+1} = n_{C_{r} } a^{r} x^{n-r} ...(i)

<u><em>Step:2</em></u>

put r = 4  

T_{4+1} = 8_{c_{4} } (x^{2} )^{4} (y^{5} )^{8-4}

T_{5} = 70 x^{8} y^{20}

<u><em>Final answer:</em></u>-

The fifth term of the binomial expansion

T_{5} = 70 x^{8} y^{20}

6 0
3 years ago
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