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Crank
3 years ago
13

Which biconditional statement is true?

Mathematics
2 answers:
tatyana61 [14]3 years ago
6 0

Answer: A shape is a rectangle if and only if the shape has exactly four sides and four right angles.

Step-by-step explanation:

  • If both a conditional statement and  its converse statement is true then we write a combine form of both the statements known as a bi-conditional statement. It is written in the if and only if form.

From the given options only option one is correct because it is true from both ways.

It can be written as If a shape has exactly four sides and four right angles then the shape is a rectangle.

  • A shape is a trapezoid if and only if the shape has a pair of parallel sides.  

It is not true because a if the shape has a pair of parallel sides then it can be a parallelogram.

  • A shape is a triangle if and only if the shape has three sides and three acute angles.

 It is not true because a if the shape has three sides and three acute angles then it is only a acute triangle.

  • A shape is a square if and only if the shape has exactly four congruent sides.

 It is not true because if the shape has exactly four congruent sides then it can also be rhombus.

So the only true bi-conditional statement is "A shape is a rectangle if and only if the shape has exactly four sides and four right angles. "

Oksana_A [137]3 years ago
4 0

Answer:

The first one: A shape is a rectangle if and only if the shape has exactly four sides and four right angles.

Step-by-step explanation:

The true statement is the first one:

A shape is a rectangle if and only if the shape has exactly four sides and four right angles.

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Find four consecutive multiples of 5 whose sum is 90
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Answer:

15, 20, 25, 30

Step-by-step explanation:

If want to find the first multiple, we could actually use the equation: x + (x+5) + (x+5+5) + (x+5+5+5) = 90. After simplifying this term, we would get 4x+30 = 90 which then transforms into 4x=60 which simplifys to x = 15. Plug 15 into the original equation, and you would get the multiples 15 + 20 + 25 + 30 = 90.

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Consider the following functions. f1(x) = x, f2(x) = x2, f3(x) = 7x − 5x2 g(x) = c1f1(x) + c2f2(x) + c3f3(x) Solve for c1, c2, a
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Answer:

Therefore the solution is = k{-7,5,1} where k ∈R

Step-by-step explanation:

Given that,

f₁(x) =x

f₂(x)= x²

f₃(x)= 7x - 5x²

Also,

g(x) = c₁f₁(x)+c₂f₂(x)+c₃f₃(x)

Putting the values of f₁(x), f₂(x) and f₃(x).

g(x) = c₁.x+c₂x²+c₃(7x-5x²)

Given condition that g(x)= 0

∴ c₁.x+c₂x²+c₃(7x-5x²)=0

⇒(c₁+7c₃)x +(c₂-5c₃)x² = 0

Comparing the coefficients of x and x²

∴c₁+7c₃=0               and       c₂-5c₃ =0

\Rightarrow c_1 =-7c_3                     \Rightarrow c_2=5c_3

Let c₃= k   [k∈R]  

Then c₁ = -7k   and   c₂=5k

Therefore the solution is = { c₁,c₂,c₃}  

                                           = {-7k, 5k, k}

                                            =k{-7,5,1}

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3 years ago
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Answer: X equals 8

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Subtract 2 from both sides of the equation

11X = 88

Divide both sides of the equation by 11

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