Which group of these squares do NOT support the Pythagorean theorem?

Mathematics

2 answers:

mrs_skeptik [129]2 years ago

6 0

C. the side lengths don’t match up

AysviL [449]2 years ago

6 0

group a because it is not a square

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The perimeter of a triangle ABC is 54. The triangle has side lengths AB=3x,BC=4x,and AC=5x. Find the length of each side.

Ilya [14]

3x+4x+5x=54
12x=54
x=4.5

Plug in the value of x into all three segments.

AB= 3*4.5= 13.5
BC= 4*4.5= 18
AC= 5*4.5= 22.5

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3 years ago

Explain WHAT WOULD HAPPEN TO motion IF friction DID NDT EXIST.

MA_775_DIABLO [31]

Answer:

If friction did not exist, it would be hard for us to slow down in running, or putting breaks on a car, and more!

If friction did not exist, your car will slide down the road even if you try your best to put the brakes on a car. Friction is a very important force to slow down things that are moving very fast

Step-by-step explanation:

Hoped this helped.

$GeniusUser$

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2 years ago

3.5 expressed in words is:

mart [117]

Yes, i believe the answer is true, have a great day :)

4 0

3 years ago

Read 2 more answers

Solve the equation for x

KengaRu [80]

You can do this 2 ways, one is knowing the special right triangles called triplets. An example of a triplet would be 3-4-5 where the largest value is the hypotenuse. In this example, the other 2 sides are 3 and 4 and the missing side is the hypotenuse, so using the knowledge of triplets, we can find that the hypotenuse is 5.

algebraically, we can use the Pythagorean theorem; a^2+b^2=c^2 where a and b are the opposite and adjacent and where c is the hypotenuse. We can then substitute in the values to get : 3^2 + 4^2 = c^2. When simplified, we get 9+16=c^2. Simplify further, and get 25=c^2. Now we Square root both sides to solve for C. Finally, we get C=5. Your missing side is 5.

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2 years ago

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. Verify y

mariarad [96]

Answer:

the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis is;

$\frac{\pi }{2} [e^2 - 1 ]$ or 10.036