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Semenov [28]
3 years ago
11

Is a triangle with sides that measure 6inches,8inches and 9inches a right triangle

Mathematics
2 answers:
dimulka [17.4K]3 years ago
7 0

Answer:

6x8x9=432 no because it equals 432 inches and a right angle is 90 inches.

Step-by-step explanation:

( : your welcome :D

AveGali [126]3 years ago
3 0
No, because, using the Pythagorean Theorum:

6^2+8^2=9^2
36+64 IS NOT = TO 81
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The rent for an office space is $674.30 per month. The office is 306 1/2 square feet in area.What is the monthly cost per square
Rufina [12.5K]
The answer is B because 674.30 divided by 306.5 equals 2.2 or $2.20
3 0
3 years ago
Solve equation 1/4=x^-1/2
Alex

Answer:

A. {16}

Step-by-step explanation:

1/4 = x^-1/2

We have the exponent rule a^-1 = 1/a^1

Applying the rule, we get

1/4 = 1/x^1/2

x^1/2 = √x

1/4 = 1/√x

Now let's cross multiplying, we get

√x  = 4

Taking square on both sides, we get

x = 16

The answer: A. {16}

Hope this will helpful.

Thank you.

3 0
3 years ago
Read 2 more answers
How many different factors does 25 have
viktelen [127]
3? i think so i don’t know if it’s right
7 0
2 years ago
Read 2 more answers
Find the distance and the midpoint of the line segment with endpoints (-1,4) and (-5,3)
Marina CMI [18]

Answer:

see explanation

Step-by-step explanation:

Calculate the distance (d) using the distance formula

d = √ (x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = (- 1,4) and (x₂, y₂ ) = (- 5, 3)

d = \sqrt{(-5+1)^2+(3-4)^2}

  = \sqrt{(-4)^2+(-1)^2}

  = \sqrt{16+1} = \sqrt{17} ≈4.12 ( to 2 dec. places )

To find the midpoint use the midpoint formula

[0.5(x₁ + x₂ ), 0.5(y₁ + y₂ ) ]

Using the same points as above then

midpoint = [0.5(- 1- 5), 0.5(4 + 3 ) ] = [0.5(- 6), 0.5(7) ] = (- 3, 3.5 )

4 0
3 years ago
If sin(A+B)=1/2 & sin(A-B)=1/3 find sin(2A)
timofeeve [1]

Answer:

sin(2A) = (2√2 + √3) / 6

Step-by-step explanation:

2A = (A+B) + (A−B)

sin(2A) = sin((A+B) + (A−B))

Angle sum formula:

sin(2A) = sin(A+B) cos(A−B) + sin(A−B) cos(A+B)

sin(2A) = 1/2 cos(A−B) + 1/3 cos(A+B)

Pythagorean identity:

sin(2A) = 1/2 √[1 − sin²(A−B)] + 1/3 √[1 − sin²(A+B)]

sin(2A) = 1/2 √(1 − 1/9) + 1/3 √(1 − 1/4)

sin(2A) = 1/2 √(8/9) + 1/3 √(3/4)

sin(2A) = 1/3 √2 + 1/6 √3

sin(2A) = (2√2 + √3) / 6

6 0
3 years ago
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