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ANEK [815]
3 years ago
8

Solve for c. -c/2+43 = 50

Mathematics
1 answer:
noname [10]3 years ago
5 0

\huge\text{Hey there!}

\mathsf{-\dfrac{c}{2}+43=50}

\large\text{SUBTRACT 43 to BOTH SIDES}

\mathsf{-\dfrac{c}{2}+43-43=50-43}

\large\textsf{CANCEL out: 43 - 43 because that gives you 0}

\large\textsf{KEEP: 50 - 43 because it helps solve for the c-value}

\mathsf{50 - 43 = \bf 7}

\large\text{NEW EQUATION: }\mathsf{-\dfrac{c}{2}=7}

\large\textsf{MULTIPLY }\mathsf{\dfrac{2}{-1}}\large\textsf{ to BOTH SIDES}

\mathsf{\dfrac{2}{-1}\times-\dfrac{c}{2}= \dfrac{2}{-1}\times7}

\large\textsf{CANCEL out: your RIGHT SIDE of the equation because that gives}\\\large\textsf{you 1}

\large\textsf{KEEP: }\mathsf{\dfrac{2}{-1}\times7}\large\textsf{ because that helps solve for the c-value}

\uparrow\large\text{SIMPLIFY THE EQUATION ABOVE \& YOU HAVE YOUR}\\\large\text{C-VALUE}\uparrow

\boxed{\boxed{\large\text{ANSWER: \huge\bf  c = -14}}}\huge\checkmark

\large\text{Good luck on your assignment and enjoy your day!}

~\frak{Amphitrite1040:)}

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Answer:

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Step-by-step explanation:

Vertices of the triangle ABC are A(-4, -2), B(1, 7) and C(8, -2).

(a). Area of the triangle ABC = \frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})] (Absolute value)

By substituting the values from the given vertices,

Area = \frac{1}{2}[(-4)(7+2)+(1)(-2+2)+8(-2-7)]

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        = (-54) unit²

Therefore, absolute value of the area = 54 square units

(b). Distance between two vertices (a, b) and (c, d)

        d = \sqrt{(a-c)^{2}+(b-d)^2}

     AB = \sqrt{(-4-1)^{2}+(-2-7)^{2}}

           = \sqrt{106}

           = 10.295 units

     BC = \sqrt{(1-8)^2+(7+2)^2}

           = \sqrt{130}

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     AC = \sqrt{(-4-8)^2+(-2+2)^2}

           = 12 units

     Perimeter of the triangle = AB + BC + AC = 10.295 + 11.402 + 12

                                                                          = 33.697

                                                                          ≈ 33.7 units

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