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DedPeter [7]
3 years ago
8

The lengths of the sides of a right triangle can be described by the equation a2 + b2 = c2, where a and b are the lengths of the

sides that form a right angle and c is the side opposite from the right angle, known as the hypotenuse. If the length of the hypotenuse of an isosceles right triangle is inches. How long, in inches, are the legs of the triangle?
Mathematics
2 answers:
Luden [163]3 years ago
5 0
An isosceles triangle contains an 2 identical sides and a hypotenuse. The hypotenuse of an isosceles triangle measures √2 as compared to its sides which is 1. In conclusion, to get the sides of an isosceles triangle when given the length of its hypotenuse, divide the length of its hypotenuse by √2 and you should be able to get the length of its sides.
Umnica [9.8K]3 years ago
4 0

Answer:

The real answer is  

A.   ±3

If you do the math your answer will come out to  

±3. then it also says that your answer needs to be in inches. the answer couldn't be anything else except for 3.

THE ANSWER IS A

get this verified

Step-by-step explanation:

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What the answer? 8x-(2x-3)=12
nevsk [136]

Answer:

x= 3/2 or 1.5

Step-by-step explanation:

First of all, you can take out the parenthesis because 8x is subtracting 2x-3.

8x-2x-3=12

6x-3=12

  +3  +3

6x= 15

6x/6= 15/6

x= 3/2 or 1.5

Hope this helps!

8 0
3 years ago
I need help with this Algebra question
Ainat [17]

Step 1 = Distributive Property ( multiplying the 2 by each term in parenthesis)

Step 2 = Addition Property ( adding like terms)

Step 3 = Addition Property (adding 34x to each side)

Step 4 = Division Property. ( dividing both sides by 17)


3 0
3 years ago
1 meter is equal to how many feet?
Fittoniya [83]
THE CORRECT ANSWER IS 3 FEET
5 0
3 years ago
Given the vectors A⃗ and B⃗ shown in the figure ((Figure 1) ), determine the magnitude of B⃗ −A⃗. A is 28 degrees above the posi
Vlad [161]

This problem is represented in the Figure below. So, we can find the components of each vector as follows:


\bullet \ cos(28^{\circ})=\frac{Adjacent}{Hypotenuse}=\frac{A_{x}}{44} \\ \\ \therefore A_{x}=44cos(28^{\circ})=38.85m \\ \\ \\ \bullet \ sin(28^{\circ})=\frac{Opposite}{Hypotenuse}=\frac{A_{y}}{44} \\ \\ \therefore A_{y}=44sin(28^{\circ})=20.65m


\bullet \ cos(56^{\circ})=\frac{Adjacent}{Hypotenuse}=\frac{-B_{x}}{26.5} \\ \\ \therefore B_{x}=-26.5cos(56^{\circ})=-14.81m \\ \\ \\ \bullet \ sin(56^{\circ})=\frac{Opposite}{Hypotenuse}=\frac{B_{y}}{26.5} \\ \\ \therefore B_{y}=26.5sin(56^{\circ})=21.97m


Therefore:

\vec{A}=(38.85, 20.65)m \\ \\ \vec{B}=(-14.81, 21.97)m


So:

\vec{B}-\vec{A}=(-14.81, 21.97)-(38.85, 20.65)=(-53.66,1.32)


Finally, the magnitude is:


\boxed{\left| \vec{B}-\vec{A}\right|=\sqrt{(-53.66)^2+(1.32)^2}=53.67m}

7 0
3 years ago
Read 2 more answers
What's the Answer to the Problem?
Elanso [62]
This would probably be B.

5 0
3 years ago
Read 2 more answers
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