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Semenov [28]
3 years ago
6

The area of a triangle is 6.75m2. If the base of the triangle is 3 m, what is the height of the triangle?

Mathematics
1 answer:
mixas84 [53]3 years ago
5 0
Answer is given above

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The point R(-3,a,-1) is the midpoint of the line segment jointing the points P(1,2,b)
wlad13 [49]

Answer:

The values are:

  • a = -5/2
  • b = -6
  • c = -7

Step-by-step explanation:

Given:

  • P = (x₁, y₁, z₁) = (1, 2, b)  
  • Q =  (x₂, y₂, z₂) = (c, -7, 4)  
  • m = R = (x, y, z) = (-3, a, -1)

To Determine:

a = ?

b = ?

c = ?

Determining the values of a, b, and c

Using the mid-point formula

m\:=\:\left(\frac{x_1+x_2}{2},\:\frac{y_1+y_2}{2},\:\frac{z_1+z_2}{2}\right)

  • As the point R(-3, a, -1) is the midpoint of the line segment jointing the points P(1,2,b)  and Q(c,-7,4), so
  • m = R = (x, y, z) = (-3, a, -1)

Using the mid-point formula

m\:=\:\left(\frac{x_1+x_2}{2},\:\frac{y_1+y_2}{2},\:\frac{z_1+z_2}{2}\right)

given

(x₁, y₁, z₁) = (1, 2, b) = P

(x₂, y₂, z₂) = (c, -7, 4) = Q

m = (x, y, z) = (-3, a, -1) = R

substituting the value of (x₁, y₁, z₁) = (1, 2, b) = P,   (x₂, y₂, z₂) = (c, -7, 4) = Q, and m = (x, y, z) = (-3, a, -1) = R in the mid-point formula

m\:=\:\left(\frac{x_1+x_2}{2},\:\frac{y_1+y_2}{2},\:\frac{z_1+z_2}{2}\right)

\left(x,\:y,\:z\right)\:=\:\left(\frac{1+c}{2},\:\frac{2+\left(-7\right)}{2},\:\frac{b+4}{2}\right)

as (x, y, z) = (-3, a, -1), so

\left(-3,\:a,\:-1\right)\:=\:\left(\frac{1+c}{2},\:\frac{2+\left(-7\right)}{2},\:\frac{b+4}{2}\right)

<u>Determining 'c'</u>

-3 = (1+c) / (2)

-3 × 2 = 1+c

1+c = -6

c = -6 - 1

c = -7

<u>Determining 'a'</u>

a = (2+(-7)) / 2

2a = 2-7

2a = -5

a = -5/2

<u>Determining 'b'</u>

-1 = (b+4) / 2

-2 = b+4

b = -2-4

b = -6

Therefore, the values are:

  • a = -5/2
  • b = -6
  • c = -7
6 0
3 years ago
Help appreciated on question in image!<br> Thanks:)
Verdich [7]

Answer:

x=-1,\:x=-7,\:x=i,\:x=-i

Step-by-step explanation:

Considering the equation

x^4+8x^3+8x^2+8x+7=0

Solving

x^4+8x^3+8x^2+8x+7

\mathrm{Factor\:}x^4+8x^3+8x^2+8x+7:\quad \left(x+1\right)\left(x+7\right)\left(x^2+1\right)

As

\mathrm{Use\:the\:rational\:root\:theorem}

a_0=7,\:\quad a_n=1

\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:7,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1

\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:7}{1}

-\frac{1}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+1

=\left(x+1\right)\frac{x^4+8x^3+8x^2+8x+7}{x+1}...[A]

Solving

\frac{x^4+8x^3+8x^2+8x+7}{x+1}

=x^3+7x^2+x+7

Putting \frac{x^4+8x^3+8x^2+8x+7}{x+1} =  x^3+7x^2+x+7 in equation [A]

So,

\left(x+1\right)\frac{x^4+8x^3+8x^2+8x+7}{x+1}...[A]

=\left(x+1\right)x^3+7x^2+x+7

As

x^3+7x^2+x+7=\left(x+7\right)\left(x^2+1\right)

So,

Equation [A] becomes

=\left(x+1\right)\left(x+7\right)\left(x^2+1\right)

So,  the polynomial equation becomes

\left(x+1\right)\left(x+7\right)\left(x^2+1\right)=0

\mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)\mathrm{Solve\:}\:x+1=0:\quad x=-1

\mathrm{Solve\:}\:x+7=0:\quad x=-7

\mathrm{Solve\:}\:x^2+1=0:\quad x=i,\:x=-i

\mathrm{The\:solutions\:are}

x=-1,\:x=-7,\:x=i,\:x=-i

Keywords: polynomial equation

Learn polynomial equation from brainly.com/question/12240569

#learnwithBrainly

5 0
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