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guajiro [1.7K]
2 years ago
11

Which conjunction is true?

Mathematics
2 answers:
My name is Ann [436]2 years ago
7 0
A is incorrect because isosceles cannot have right angles and right triangles cannot be isosceles. A square is not a rhombus, so B is incorrect. And yes, a kite is a rhombus and a parallelogram. You didn't finish C, but right triangles cannot be isosceles, but right triangles can have two acute angles 
Alik [6]2 years ago
5 0

Answer explanation:

1 . The given conjunction is : An isosceles triangle has two congruent sides and a right angle.

It can be or cannot be true.  An isosceles triangle can have one right angle and two other angles equal to each other , each being 45°.But it is a special isosceles triangle called ,isosceles Right triangle.

So, this conjunction is not  completely true.

→2. The Given conjunction is : A square is a: Rhombus and a rectangle.

Yes , this conjunction is true, Because ,Square Satisfies all the properties of Rhombus and Rectangle.

The Properties of Square are

1. All sides equal and each interior angle equal to 90°.

2. Opposite sides are parallel.

3. Diagonals bisect each other at right angles.

Properties of Rhombus

1. All sides are equal.

2. Opposite sides are parallel.

3. Diagonals bisect each other at right angles.

Properties of Rectangle

1.Opposite sides are equal.

2. Diagonals bisect each other.

3.Opposite sides are parallel.

4. Each interior angle measures 90°.

⇒Square Satisfies all the properties of Square and rhombus.

Yes, this Conjunction is true.

→→3rd and 4th Conjunction is incorrect, which is→ A kite is rhombus and a parallelogram,is incorrect because kite has,two pair of adjacent sides which are equal, but a rhombus has all sides  equal,whereas parallelogram has opposite sides equal.

→ A right triangle ,is isosceles and has two acute angles.→A right  triangle is isosceles,is a special case, otherwise it has two acute angles and one right angle is true statement,but it is isosceles is not true.

Option B, is a true Conjunction:→ A square is a b.rhombus and a rectangle.

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Calculate the sum of the first 36 terms of the arithmetic sequence defined in which a36=14 and the common difference is d=1/8
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Answer:

425.25


Step-by-step explanation:

Since we are given 36th term as 14 and we know common difference is \frac{1}{8}, it means that from the first term, we add \frac{1}{8} to each and get 14 on the 36th term. To figure out the first term, thus, we have to subtract \frac{1}{8} 35 times from 14. Let's do it to get first term:

14-35(\frac{1}{8})=\frac{77}{8}=9.625

The sum of arithmetic sequence formula is:

S_{n}=\frac{n}{2}[2a+(n-1)d]

Where,

  • S_{n} is the sum of nth term (we want to figure this out for first 36 terms)
  • a is the first term (we figured this out to be 9.625)
  • n is the term number (36 for our case)
  • d is the common difference (given as \frac{1}{8})

Substituting all the values, we get:

S_{36}=\frac{36}{2}[2(9.625)+(36-1)(\frac{1}{8})]\\S_{36}=18[19.25+4.375]\\S_{36}=18[23.625]\\S_{36}=425.25

First answer choice is right.


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