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krok68 [10]
3 years ago
7

Ast year, 3,120 students competed in a statewide athletic competition. If each participating school district sent 26 students to

the competition, how many participating school districts were there?
Mathematics
2 answers:
andreyandreev [35.5K]3 years ago
3 0

Answer: 120 schools

Step-by-step explanation:

3,120 students competed in a statewide athletic competition and each participating school district sent 26 students to the athletic competition.

Total number of students= 3120

Total students sent by each school= 26

Number of participating school districts= Total number of students ÷ he students sent by each school

= 3120 / 26

= 120 students

There are 120 participating schools in the competition

velikii [3]3 years ago
3 0

Answer:

There were 120 school districts participating in the competition.

Step-by-step explanation:

In order to calculate the amount of school districts that participated in the competition we need to take the total amount of students that were present and divide it by the number of students each district sent. In this case there were 3,120 students and each district sent 26 students. So we have:

number of districts = 3120/26

number of districts = 120

There were 120 school districts participating in the competition.

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How many different right triangles are there with a hypotenuse of lenght 5 cm
Sveta_85 [38]

Answer:

Draw a horizontal line across a page of paper, somewhere in the middle. Mark a point A on the line, toward the left margin.

Spread open a compass to make good sized circle, but so that if the point of the compass is on the point you drew, the pencil fits on the page, both below the top edge and to the left of the right edge of the page. Draw the arc, from roughly the 12 O’Clock position over and down to the intersection of the line segment, at the 3 O’Clock position.

Call the opening of your compass, the radius of the arc you just drew, “5 units”.

Pick any point on the arc between the 12 and 3 positions, B. Drop a line down from that point B, perpendicular to the original horizontal line. Label the point that it intersects the horizontal line, C.

ABC is a right triangle with hypotenuse 5. Do this again with a point a little closer to the 3 O’Clock position. It’s another right triangle with hypotenuse 5.

Indeed, as you get closer to the right, along the arc, the height of the triangle declines, but the width of the triangle increases. The hypotenuse remains 5 in all cases. We picked points on the circle. We could have picked points on the horizontal line first, and raised perpendicular lines until they intersected the circle. Each point forms a distinct right triangle.

There are as many possible right triangles as there are possible points in a line segment.

Step-by-step explanation:

6 0
3 years ago
AB:BC is 3:4. Solve for x.
Sphinxa [80]
I got B but I'll show you how I got that answer.

Let's look at AB and BC. AB = 3 and BC = 4.
3/4 = 0.75

On the triangle, AB = 90. And BC = 10x - 20. We need to solve for x and prove that AB/BC = 0.75.

Let's start with Choice A.

10(11) - 20 = 90.

AB = 90
BC = 90.

90/90 = 1. It doesn't equal 0.75. Meaning this answer is wrong.

Now let's look at Choice B.

10(14) - 20 = 120.

AB = 90
BC = 120

90/120 = 0.75

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6 0
3 years ago
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3 years ago
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tankabanditka [31]
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8 0
3 years ago
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PLEASE HELP AND SHOW WORK IF YOU CAN​
Murljashka [212]

Answer:

c) XY > RS

= true

Step-by-step explanation:

Firstly know the meaning of each sign given

> means <u>is greater than</u>

= means <u>is equal to</u>

a)XZ > ST

XZ = 8

ST = 11 - greater

= false

b) XY = RS

Due to the angles presented, we conclude that the greater the angle, the longer the unmeasured line.

XY is longer than RS

= false

c) XY > RS

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d) RS > XY

RS is shorter than XY hence not greater.

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3 years ago
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