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Anvisha [2.4K]
3 years ago
6

Solve for x: -10x/2+x=x+8/x

Mathematics
1 answer:
rosijanka [135]3 years ago
3 0

The\ domain:\\D:\ x\neq-2\ \wedge\ x\neq0\\\\\dfrac{-10x}{2+x}=\dfrac{x+8}{x}\ \ \ \ |\text{cross multiply}\\\\(-10x)(x)=(2+x)(x+8)\\\\-10x^2=(2)(x)+(2)(8)+(x)(x)+(x)(8)\\\\-10x^2=2x+16+x^2+8x\\\\-10x^2=x^2+10x+16\ \ \ +10x^2\\\\11x^2+10x+16=0\\\\a=11,\ b=10,\ c=16\\\\b^2-4ac=10^2-4(11)(16)=100-704=-604 < 0\\\\NO\ REAL\ SOLUTION

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Answer:

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Step-by-step explanation:

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What is the measure of angle c?​
s344n2d4d5 [400]

Answer:

50 degrees

Step-by-step explanation:

Okay! So first off, because this is NOT a right triangle, we can't use soh cah toa. That means we can either use the law of sines or the law of cosines.

Because we only have sides here and no angles, we are forced to use the law of cosines.

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3 years ago
The expected activity time in PERT analysis is calculated as: the weighted average of a, m, and b, with m weighted 4 times as he
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Answer:

\text{the weighted average of a, m, and b, with m weighted 4 times as heavily as a and b.}

Step-by-step explanation:

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2 years ago
the length of a rectagle is 5 in longer than its width. if the perimeter of the rectangle is 58 in, find its length and width
dolphi86 [110]

Answer:

  • Length = 17 inches

  • Width = 12 inches

⠀

Step-by-step explanation:

⠀

As it is given that, the length of a rectangle is 5 in longer than its width and the perimeter of the rectangle is 58 in and we are to find the length and width of the rectangle. So,

⠀

Let us assume the width of the rectangle as x inches and therefore, the length will be (x + 5) inches .

⠀

Now, <u>According to the Question :</u>

⠀

{\longrightarrow \qquad { \pmb{\frak {2 ( Length + Breadth )= Perimeter_{(Rectangle)} }}}}

⠀

{\longrightarrow \qquad { {\sf{2 ( x + 5 + x )= 58 }}}}

⠀

{\longrightarrow \qquad { {\sf{2 ( 2x + 5  )= 58 }}}}

⠀

{\longrightarrow \qquad { {\sf{ 4x + 10= 58 }}}}

⠀

{\longrightarrow \qquad { {\sf{ 4x = 58  - 10}}}}

⠀

{\longrightarrow \qquad { {\sf{ 4x = 48}}}}

⠀

{\longrightarrow \qquad { {\sf{ x =  \dfrac{48}{4} }}}}

⠀

{\longrightarrow \qquad{ \underline{ \boxed { \pmb{\mathfrak {x = 12}} }}} }\:  \:  \bigstar

⠀

Therefore,

  • The width of the rectangle is 12 inches .

⠀

Now, We are to find the length of the rectangle:

{\longrightarrow \qquad{ { \frak{\pmb{Length = x + 5 }}}}}

⠀

{\longrightarrow \qquad{ { \frak{\pmb{Length = 12 + 5 }}}}}

⠀

{\longrightarrow \qquad{ { \frak{\pmb{Length = 17}}}}}

⠀

Therefore,

  • The length of the rectangle is 17 inches .

⠀

8 0
2 years ago
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