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Eva8 [605]
3 years ago
13

A city park designer is designing a new park. The park will be shaped like a right triangle and there will be two pathways for p

edestrians,shown by VT and VW in the diagram. The park planner only wrote two lengths on his sketch as shown.Based on the diagram,what will be the lengths of the two pathways?
The length of pathway VW is _ yards.

The length of pathway VT is _ yards.

Mathematics
1 answer:
Aloiza [94]3 years ago
8 0

Answer:

VW=36

VT=27

Step-by-step explanation:

VW is the midsegment of the triangle

so

VW=72/2

=36

VXT is a right angle triangle

so

a²+b²=c²

a=√c²-b²

a=√45²-36²

a=√(45-36)(45+36)

a=√9×81

a=√9×√81

a=27

therefore

VT=27

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4 years ago
Please help me for the brainliest answer
Oxana [17]

Answer:

aₙ= -2n²

Step-by-step explanation:

<h2><u>Solution 1:</u></h2>

The sequence:

  • -2, -8, -18, -32, -50

The difference between the terms:

  • -6, -10, -14, -18
  • a₁= -2
  • a₂= a₁ - 6 = a₁ - 2*3= a₁- 2*(2²-1)
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  • a₄= a₃- 14= a₁ - 30= a₁ - 2*15= a₁ - 2*(4² -1)
  • ...
  • aₙ= a₁ -2*(n²-1)= -2 -2n² +2= -2n²

As per above, the nth term is: aₙ= -2n²

<h2><u /></h2><h2><u>Solution 2</u></h2>

The sequence:

  • -2, -8, -18, -32, -50
  • -2*1, -2*4, - 2*9, -2*25
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6 0
3 years ago
The difference between the two roots of the equation 3x^2+10x+c=0 is 4 2/3 . Find the solutions for the equation.
andrezito [222]

Answer:

Given the equation: 3x^2+10x+c =0

A quadratic equation is in the form: ax^2+bx+c = 0 where a, b ,c are the coefficient and a≠0 then the solution is given by :

x_{1,2} = \frac{-b\pm \sqrt{b^2-4ac}}{2a} ......[1]

On comparing with given equation we get;

a =3 , b = 10

then, substitute these in equation [1] to solve for c;

x_{1,2} = \frac{-10\pm \sqrt{10^2-4\cdot 3 \cdot c}}{2 \cdot 3}

Simplify:

x_{1,2} = \frac{-10\pm \sqrt{100- 12c}}{6}

Also, it is given that the difference of two roots of the given equation is 4\frac{2}{3} = \frac{14}{3}

i.e,

x_1 -x_2 = \frac{14}{3}

Here,

x_1 = \frac{-10 + \sqrt{100- 12c}}{6} ,     ......[2]

x_2= \frac{-10 - \sqrt{100- 12c}}{6}       .....[3]

then;

\frac{-10 + \sqrt{100- 12c}}{6} - (\frac{-10 + \sqrt{100- 12c}}{6}) = \frac{14}{3}

simplify:

\frac{2 \sqrt{100- 12c} }{6} = \frac{14}{3}

or

\sqrt{100- 12c} = 14

Squaring both sides we get;

100-12c = 196

Subtract 100 from both sides, we get

100-12c -100= 196-100

Simplify:

-12c = -96

Divide both sides by -12 we get;

c = 8

Substitute the value of c in equation [2] and [3]; to solve x_1 , x_2

x_1 = \frac{-10 + \sqrt{100- 12\cdot 8}}{6}

or

x_1 = \frac{-10 + \sqrt{100- 96}}{6} or

x_1 = \frac{-10 + \sqrt{4}}{6}

Simplify:

x_1 = \frac{-4}{3}

Now, to solve for x_2 ;

x_2 = \frac{-10 - \sqrt{100- 12\cdot 8}}{6}

or

x_2 = \frac{-10 - \sqrt{100- 96}}{6} or

x_2 = \frac{-10 - \sqrt{4}}{6}

Simplify:

x_2 = -2

therefore, the solution for the given equation is: -\frac{4}{3} and -2.


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