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Alex_Xolod [135]
2 years ago
14

What value of w would make the equation true

Mathematics
1 answer:
Rashid [163]2 years ago
6 0

Step-by-step explanation:

\frac{12w + 8}{6}  =  \frac{8w - 16}{3 }  \\ 36w + 24 = 48w - 96 \\ 36w - 48w =  - 96 - 24 \\  - 12w =  - 120 \\ w = 10

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Ab with slope = 5 and ST with slope = 5
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Parallel

Step-by-step explanation:

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2 years ago
Plase dont awnser just for points
egoroff_w [7]
So 3 2/3 is about 3.67 so it would go in between 3.65 and 3.7. So it would be the second answer.
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3 years ago
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Given that f(x) = x2 + 6x – 2, g(x) = x – 7, and h(x) = x + 4 find each function.
choli [55]

Answer:

A) x² + 7x - 9

Step-by-step explanation:

Deduct\Add each like-term to arrive at your answer.

3 0
3 years ago
) the top and bottom margins of a poster are 4 cm and the side margins are each 5 cm. if the area of printed material on the pos
grin007 [14]
If x represents the width of the poster (including borders), the area of the finished poster can be written as
.. a = x*(390/(x -10) +8)
.. = 8x +390 +3900/(x -10)

Then the derivative with respect to x is
.. da/dx = 8 -3900/(x -10)^2
This is zero at the minimum area, where
.. x = √(3900/8) +10 ≈ 32.08 . . . . cm
The height is then
.. 390/(x -10) +8 = 8 +2√78 ≈ 25.66 . . . . cm

The poster with the smallest area is 32.08 cm wide by 25.66 cm tall.

_____
In these "border" problems, the smallest area will have the same overall dimension ratio that the borders have. Here, the poster is 10/8 = 1.25 times as wide as it is high.
8 0
3 years ago
Please please please help and solve this with steps, much help needed thank you :) 20 points for this!
leonid [27]

Answer:

y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}

(Please vote me Brainliest if this helped!)

Step-by-step explanation:

\frac{dy}{dx}y=x^3+2x-5x+3

\mathrm{First\:order\:separable\:Ordinary\:Differential\:Equation}

  • \mathrm{A\:first\:order\:separable\:ODE\:has\:the\:form\:of}\:N\left(y\right)\cdot y'=M\left(x\right)

1. \mathrm{Substitute\quad }\frac{dy}{dx}\mathrm{\:with\:}y'\:

y'\:y=x^3+2x-5x+3

2. \mathrm{Rewrite\:in\:the\:form\:of\:a\:first\:order\:separable\:ODE}

yy'\:=x^3-3x+3

  • N\left(y\right)\cdot y'\:=M\left(x\right)
  • N\left(y\right)=y,\:\quad M\left(x\right)=x^3-3x+3

3. \mathrm{Solve\:}\:yy'\:=x^3-3x+3:\quad \frac{y^2}{2}=\frac{x^4}{4}-\frac{3x^2}{2}+3x+c_1

4. \mathrm{Isolate}\:y:\quad y=\sqrt{\frac{x^4-6x^2+12x+4c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+4c_1}{2}}

5. \mathrm{Simplify}

y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}

7 0
3 years ago
Read 2 more answers
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