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insens350 [35]
3 years ago
15

PLEASE HELP DUE SOON! WILL GIVE BRAINLIEST AND POINTS!

Mathematics
2 answers:
Fynjy0 [20]3 years ago
3 0

3.  6/8/10 is twice the common Pythagorean Triple 3/4/5.

Answer: A.   10

4. 3/4/5 is a Pythagorean Triple because they're three natural numbers that satisfy a²+b²=c^2, here 3²+4²=5².

Answer: C

Non-zero whole numbers are called <em>natural numbers.</em>

<em />

<em />

Aleksandr-060686 [28]3 years ago
3 0

Answer:

3  10

4  C yes it is a pythagorean triple 3^2 +4^2 = 5^2

Step-by-step explanation:

We can find the unknown side by using the Pythagorean theorem

a^2 +b^2 = c^2

6^2 +8^2 = c^2

36+64 =c^2

100 = c^2

Take the square root of each side

sqrt(100) = sqrt(c^2)

10 = c

4. a^2 +b^2 = c^2

   3^2 +4^2 = 5^2

   9+16= 5^2

    25 = 25

Since the numbers fit the Pythagorean theorem, it is a Pythagorean triple

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aleksklad [387]

Answer:

a.2nd quarter with 9 goals

b. 4.8 goals

c. 4 goals

Step-by-step explanation:

a. The mode is defined as the most appearing data point or the data point with the highest frequency..

From our data(for away goals):

  • 1st quarter-2
  • 2nd quarter-9
  • 3rd quarter-7
  • 4th quarter-4

Hence, the 2nd quarter has the mode for away goals with 9 goals.

b. Mean is defined as the average of a set of data points.

#We calculate the totals goals per quarter, sum over all quarters then divide by the number of games, 10:

\bar x=\frac{1}{n}\sum{x_i}\\1^{st }_g=Away+Home=5+2=7\\\\2^{nd}_g=Away+Home=4+9=13\\\\3^{rd}_g=Away+Home=8+7=15\\\\4^{th}_g=Away+Home=9+4=13\\\\\bar x=\frac{1}{n}\sum{x_i}=\frac{1}{10}(7+13+15+13)=4.8

Hence, the mean number of goals per quarter is 4.8 goals

c. To find the number of more home goals than away goals, we subtract from their summations as:

g_m=\sum{g_h}-\sum{g_a}\\\\=(5+4+8+9)-(2+9+7+4)\\\\=26-22\\\\=4

Hence, there are 4 more home goals than away goals.

4 0
3 years ago
In the circle below, suppose m FEH=272º and m EFG=116º. Find the following.​
adoni [48]

Answer:

m∠FEH = 44°

m∠EHG =  64°

Step-by-step explanation:

1) The given information are;

The angle of arc m∠FEH = 272°, the measured angle of ∠EFG = 116°

Given that m∠FEH = 272°, therefore, arc ∠HGF = 360 - 272 = 88°

Therefore, angle subtended by arc ∠HGF at the center = 88°

The angle subtended by arc ∠HGF at the circumference = m∠FEH

∴ m∠FEH = 88°/2 = 44° (Angle subtended at the center = 2×angle subtended at the circumference)

m∠FEH = 44°

2) Similarly, m∠HGF is subtended by arc m FEH, therefore, m∠HGF = (arc m FEH)/2 = 272°/2 = 136°

The sum of angles in a quadrilateral = 360°

Therefore;

m∠FEH + m∠HGF + m∠EFG + m∠EHG = 360° (The sum of angles in a quadrilateral EFGH)

m∠EHG = 360° - (m∠FEH + m∠HGF + m∠EFG) = 360 - (44 + 136 + 116) = 64°

m∠EHG =  64°.

5 0
3 years ago
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