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Norma-Jean [14]
2 years ago
7

7. All of the following are perfect square trinomials EXCEPT one. Which is it?

Mathematics
1 answer:
wolverine [178]2 years ago
8 0

Answer:

All answers EXCEPT answer C. are perfect square trinomials.

Step-by-step explanation:

A perfect square trinomial is polynomial that satisfies the following condition:

(a+b)^{2} = a^{2}+2\cdot a \cdot b + b^{2}, \forall\,a,b\in\mathbb{R}

Let prove if each option observe this:

a) 4\cdot x^{2} + 12\cdot x\cdot y + 9\cdot y^{2}

1) 4\cdot x^{2} + 12\cdot x\cdot y + 9\cdot y^{2} Given

2) (2\cdot x)^{2}+ 2\cdot (2\cdot x)\cdot (3\cdot y)+(3\cdot y)^{2} Definition of power/Distributive, associative and commutative properties.

3) a = 2\cdot x, b = 3\cdot y Definition of perfect square trinomial/Result.

b) 9\cdot a^{2}-36\cdot a + 36

1) 9\cdot a^{2}-36\cdot a + 36 Given.

2) (3\cdot a)^{2}-2\cdot (3\cdot a)\cdot 6 + 6^{2} Definition of power/Distributive, associative and commutative properties.

3) a = 3\cdot a, b = 6 Definition of perfect square trinomial/Result.

c) x\cdot y^{2}-4\cdot x^{2}\cdot y^{2}+4\cdot x^{2}\cdot y^{2}

1) x\cdot y^{2}-4\cdot x^{2}\cdot y^{2}+4\cdot x^{2}\cdot y^{2} Given

2) x\cdot y\cdot (1-4\cdot x+4\cdot x) Distributive property.

3) x\cdot y \cdot 1 Existence of the additive inverse/Modulative property.

4) x\cdot y Modulative property/Result.

d) a^{2}\cdot b^{2}+4\cdot a^{3}\cdot b + 4\cdot a^{4}

1) a^{2}\cdot b^{2}+4\cdot a^{3}\cdot b + 4\cdot a^{4} Given

2) (a\cdot b)^{2}+2\cdot (a\cdot b)\cdot (2\cdot a^{2})+(2\cdot a^{2}) Definition of power/Distributive, associative and commutative properties.

3) a = a\cdot b, b = 2\cdot a^{2} Definition of perfect square trinomial.

All answers EXCEPT answer C. are perfect square trinomials.

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Answer:

a) x  =  0,70 miles

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b) V(min)  = 2.59 m/hr

Step-by-step explanation:

Let assume the boat is in Point A,  she lands in point B, and R  is the restaurant

Let call  x distance between the point O in  which perpendicular line from the boat get to shoreline, and the point were she land.

We know  distance is

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She rows at 2 miles/hr  and walk at 3 miles /hr

According to that she takes

c (hypotenuse) of right triangle  AOB/ 2 rowing

and  15 - x /3  walking

Total time is:

T  =  c/2  + ( 15-x )/3          c =√(2)² + x²     ⇒  T = [√(2)² + x²  ] /2  +  ( 15-x )/3  

T  =  1/2* (√4 + x²)  +( 15 - x )/3   (1)

And that is  the Objective function to minimize

a) Taking derivatives on both sides of the equation we get

T´(t)  =  x /( √4 + x²)  - 1/3    ⇒   T´(t)  = 0     x /( √4 + x²)  - 1/3 = 0

3*x - √( √4 + x²) = 0     ⇒  3*x  =  √( √4 + x²)

Squaring both sides

9x²  =  4 + x²     ⇒  8x²  = 4       x² = 1/2      x  =  0,70 miles

If we plug this value in the Objective function we will get the minimum time

1/2* (√4 + x²)  +( 15 - x )/3    ⇒ [√ 4  + 0,5] /2 +  14,3/3 = T (min)

T(min)  =  1.06 + 4.77  = 5.83 hr

Distance L (hypotenuse of right triangle AOR)

L = √(2)²  + (15)²   L  = 15,13 miles

And that distance have to be traveled at least in 5.83 hr rowing

Then  as  v = d/t      V(min)  =  15.13/ 5.83   ⇒    V(min)  = 2.59 m/hr

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2 years ago
A. Show one pair of corresponding points and two pairs of corresponding sides in the original polygon and its copy. B. what scal
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Answer:

Step-by-step explanation:

(A).Two corresponding sides of the given figures will be,

AB and DE

BC and EF

Similarly one pair of corresponding points will be,

A and D.

(B). Since large figure has been scaled to the smaller image,

Scale factor = \frac{\text{Side of the image}}{\text{Side of the actual figure}}

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Therefore, scale factor will be 0.25.

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Answer:

<A = 90°, <B = 71.6° <C = 18.4°

Step-by-step explanation:

Given the vertices of a △ABC to be A(−1, 6), B(2, 10), and C(7, −2)

Before we can get each angle of the triangle, we need to get its sides first.

Using the formula fie finding the distance between two points

D = √(x2-x1)²+(y2-y1)²

For side AB:

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AC = √{7-(-1)}²+(-2-6)²

AC = √8²+8²

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AC = √64×2

AC = 8√2

For side BC:

Given B(2,10) and C(7, −2)

BC = √(7-2)²+(-2-10)²

BC = √5²+12²

BC = √25+144

BC = √169

BC = 13

First we need to get theta using cosine rule

|AC|² = |AB|+|BC| - 2|AB||AC| cos theta

(8√2)² = 5²+13²-2(5)(13)cos theta

128 = 169-130costheta

128-169 = -130costheta

-41= -130costheta

Costheta = -41/-130

Cos theta = 0.315

theta = arccos 0.315

theta = 71.64°

<B = 71.6°

Using sine rule to get <A,

BC/sin A = AC/sinB

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<C = 180°-(71.6+90)

< C = 180-161.6

<C = 18.4°

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