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podryga [215]
2 years ago
10

Are lines e and f parallel? Show your work.

Mathematics
1 answer:
SashulF [63]2 years ago
5 0

Answer:

last two both should be = 180

Step-by-step explanation:

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29. Given the information regarding AABC and ADEF,
ziro4ka [17]

Answer:

B

Step-by-step explanation:

just because :)

4 0
2 years ago
You estimate that there are 75 marbles in a jar. The actual amount is 82 marbles. Find the percent error
Vikentia [17]

Answer:

Percent Error:: 12/36 = 33 1/3 %

Step-by-step explanation:

5 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
What is the equation of the graphed line?
labwork [276]
Y=.25x is the only resonable answer

3 0
3 years ago
Which equation has the solutions x = 1+
BaLLatris [955]

Answer:

The last equation x2 - 2x -4 = 0

has solution  (x - 1)^2 - 5 = 0,   x  =  1 + root(5)  or  x = 1 - root(5)

Step-by-step explanation:

If a quadratic function has roots 1 and 5

f(x) = (x -1)(x- 5)

f(x) = x^2 - 6x + 5

Unless you meant.  -4 and 6  ?

g(x) = (x + 4)(x - 6)

g(x) = x^2 -2x -24

-------------------------

Or did you mean  x = 1 and x =4 ?...

x^2 + 2x + 4 = 0  :   complete square  x^2 + 2x + 1 + 3 = 0,   (x+1)^2 + 3 = 0

x^2 - 2x + 4 = 0 :  complete square:  (x -1)^2 + 3 = 0

0x^2 + 2x - 4 = 0,    2x - 4 = 0,  x = 2

x^2 - 2x - 4 = 0  becomes:   x^2 - 2x + 1 - 1 -4 = 0 ;  (x - 1)^2 - 5 = 0

5 0
3 years ago
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