WILL GIVE BRAINLIEST!!How do identify which side is the hypotenuse of a right triangle?

1 answer:

8 0

The hypotenuse of a right triangle is always the side opposite the right angle. it is the longest side in a right triangle. the other two sides are called the opposite and adjacent sides. these sides are labeled in relation to an angle

You might be interested in

The following three shapes are based only on squares, semicircles, and quarter circles. Find the perimeter and the area of each

bixtya [17]

When we divide the figure in four parts, we obtain four squares with its sides of 5 centimeters of lenght, with quarter circles (two) in each one of them. The limits of the shadded area are arcs of cirncunference

To calculate the area of the shadded part, we must choose one square of 5cmx5cm and divide it in 3 sectors:

a: the area below the shaded area.
b: the area above the shaded area.
c: the shaded area.

The area of the square (As) is:
As=L²
As=(5 cm)²
As=25 cm²

The area of a circunference is: A=πR² (R:radio), but we want the area of the quarter circle, so we must use A=1/4(πR²), to calculate the area of the sectors a+c:
A(a+c)=1/4(π(5)²)
A(a+c)=25π/4

The area of the sector "b" is:
Ab=As-A(a+c)
Ab=25-25π/4
Ab=25(1-π/4)

The area of the sector a+b, is:
A(a+b)=2Ab
A(a+b)=2x25(1-π/4)
A(a+b)=50(1-π/4)

Then, the shadded area (Sector c) is:
Ac=As-A(a+b)
Ac=25-50(1-π/4)
Ac=25-(50-50π/4)
Ac=25-50+50π/4
Ac=(50π/4)-25
Ac=(25π/2)-25
Ac=25(π/2-1)

The area of each shaded part is: 25(π/2-1)

To calculate the perimeter of a shaded part, we must remember that the perimeter of a circunference is: P=2πR. If we want the perimeter of a quarter circle we must use: P= 2πR/4. But there is two quarter circles in the square of 5cmx5cm, so the perimeter of the shaded area is:

P=2(2πR/4)
P=4πR/4
P=πR
P=5π

The perimeter of each shaded part is: 5π

6 0

3 years ago

Can someone please Find the Y, the equation to find it is in the top middle. I need the Y found for all 5 X points

katovenus [111]

Hope this helps you! :)

4 0

3 years ago

Jessie works at least 36 hours a week. She works as a math

Olegator [25]

Answer:

The feasible region is unbounded which is the shaded portion in the graph.

Step-by-step explanation:

Given that, the number of hours Jessie works as a teacher and tutor is y and x respectively, which can't be negative.

So, $x\ge0\;\cdots(i)$

y\ge0\;\cdots(ii)

She works at least 36 hours a week.

$\Rightarrow x+y\ge 36$

$\Rightarrow y\ge 36-x\;\cdots (iii)$

She works as a teacher for al least twice the number of hours she works as a tutor.

$\Rightarrow y\ge2x\;\cdots(iv)$

The boundary of the inequalities from equations (i), (ii), (iii), and (iv) are

$x=0, y=0, y=36-x$ and $y=2x$ .

The feasible region is the shaded portion which is the unbounded region as shown in the graph. Any point P(x,y) from the feasible region will satisfies all the constraints.