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Kaylis [27]
3 years ago
10

Please help this is urgent!

Mathematics
2 answers:
Diano4ka-milaya [45]3 years ago
4 0

Answer:

Isosceles

Obtuse

Step-by-step explanation:

1) When two sides of a triangle are the same length, the triangle is an isosceles triangle.

2) When one angle of the triangle is greater than 90 degrees, the triangle is an obtuse triangle.

I hope this helps! Have a great day!

Elanso [62]3 years ago
4 0

Answer:

Isosceles, obtuse

Step-by-step explanation:

There are <em>three</em> types of triangles based on their sides:

  1. <u>Equilateral</u>: a triangle with 3 equal sides
  2. <u>Isosceles</u>: a triangle with 2 equal sides
  3. <u>Scalene</u>: a triangle with no equal sides

This triangle here as sides of 28 cm, 16 cm, and 16 cm

  • This triangle has two equal sides of 16 cm, indicating it is an isosceles triangle

There are <em>three</em> types of triangles based on their angles:

  1. <u>Acute</u>: when all angles are less than 90°
  2. <u>Right</u>: when the triangle has one angle that is 90°
  3. <u>Obtuse</u>: when one of the angles is greater than 90°

This triangle has angles of 26°, 26°, and 128°

  • This triangle has one angle that is greater than 90° → 128°, indicating that this is an obtuse triangle
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find the product with the exponent in the simplest form. Then, identify the values of x and y. 6^1/3 * 6^1/4 = 6^x/y
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The key to solving this problem is understanding properties of exponents:
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6 0
3 years ago
5(7 + 3x) - 12 how to simplify​
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To solve this we need to use this formula:

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6 0
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sammy [17]

Answer:

Confidence Interval: (21596,46428)

Step-by-step explanation:

We are given the following data set:

10520, 56910, 52454, 17902, 25914, 56607, 21861, 25039, 25983, 46929

Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{340119}{10} = 34011.9

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Confidence interval:

\mu \pm t_{critical}\frac{\sigma}{\sqrt{n}}

Putting the values, we get,

t_{critical}\text{ at degree of freedom 9 and}~\alpha_{0.05} = \pm 2.2621

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