In a 30-60-90 triangle the long leg is half the hypotenuse.

2 answers:

7 0

And because we know that we cut the base of the equilateral triangle in half, we can see that the side opposite the 30° angle (the shortest side) of each of our 30-60-90 triangles is exactly half the length of the hypotenuse.

Masja [62]2 years ago

4 0

Answer:

Never

Step-by-step explanation:

In a 30°−60°−90° triangle, the length of the hypotenuse is twice the length of the shorter leg, and the length of the longer leg is √3 times the length of the shorter leg. To see why this is so, note that by the Converse of the Pythagorean Theorem, these values make the triangle a right triangle.

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HELP PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!! CERTAIN ANSWERS AND I WILL MARK YOU AS BRAINIEST!!!! WISH YOU BEST OF LUCK HOPEFULLY ITS N

AURORKA [14]

Answer:

Minimum is a 4th Degree

Positive; Even

$x=-4, -1\text{ Odd Multiplicity}\\x=3\text{ Even Multiplicity}$

Step-by-step explanation:

Part 1)

The minimum degree of our function will be 4.

Looking at the graph, we know that the graph crosses the x-axis at -4 and -1. Since it crosses through the x-axis at these two points, these two factors must have an odd multiplicity.

So, it can be anything 1, 3, 5, 7, etc.

However, we will choose the lowest one, 1.

Next, we know that the graph bounces off at 3.

So, it must have an even multiplicity. In other words, 2, 4, 6, 8, etc.

We choose the lowest one, 2.

Therefore, the minimum degree of our function will be 1+1+2 or 4.

Part 2)

The degree of our polynomial is (and will always be) even. Therefore, both ends of the graph will go in the same direction.

Recall the simplest even polynomial, the parent quadratic function. When the leading coefficient is positive, both of the ends go straight up.

This applies to all polynomials with even degrees.

Therefore, since the arms of the graph is going straight up towards positive infinity, the leading coefficient of our graph must be positive.

Part 3)

This is similar to Part 1.

We can see that the graph touches the x-axis at -4, -3, and 1. So, the zeros of the function is: $x=-4, -1, 3$

We know that it passes through $x=-4 \text{ and } x=-1$ . So, these two factors must have an odd multiplicity.

However, since the graph bounces off $x=3$ , this factor must have an even multiplicity.

And we're done!

7 0

3 years ago

A student ran out of time on a multiple choice exam and randomly guessed the answers for two problems. Each problem had 5 answer

choli [55]

Answer:

there is a 64% chance that the student got both problems wrong

a 32% chance that they got only 1 correct

and a 4% chance that they got both correct

Step-by-step explanation:

There are 25 total possible combinations of answers, with 8 possible combinations where the student would get 1 answer right, and 1 combination where the student would get both answers correct.

$25-9=16$