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antoniya [11.8K]
3 years ago
11

I'm justing starting geometry honors, and I would like to know how to pass this class?! Is it easy or hard? HELP!!

Mathematics
2 answers:
rosijanka [135]3 years ago
8 0
Its different for different people but for me it was easy.
a_sh-v [17]3 years ago
5 0
I found geometry to be very easy it is a lot of fun brain teasers if u CAN do logic puzzles u will do well and enjoy it if you have a good teacher
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Let U = {1, 2, 3, 4, 5, 6, 7}, A= {1, 3, 4, 6}, and B= {3, 5, 6}. Find the set A’ U B’
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Answer:

Step-by-step explanation:

A'={2,5,7}

B'={1,2,4,7}

A'UB'={1,2,4,5,7}

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3 years ago
Write the equation of a circle with center (6,4) that passes through the coordinate (2,1). In your final answer, include all of
Elodia [21]
All points along the circle with be the distance of the radius from the center...so the radius can be found using the Pythagorean Theorem..

r^2=(4-1)^2+(6-2)^2

r^2=9+16

r^2=25

r=5

The equation of the circle can be expressed as:

r^2=(x-h)^2+(y-k)^2 where (h,k) correspond to the center of the circle, (2,1) in this case.

(x-2)^2+(y-1)^2=25

if you wanted it in a more standard form...

(y-1)^2=25-(x-2)^2

(y-1)^2=25-x^2+4x-4

(y-1)^2=-x^2+4x+21

y-1=(-x^2+4x+21)^(1/2)

y=1(+/-)(-x^2+4x+21)^(1/2)
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3 years ago
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Using identities, find a)(2 + 3) 2 b) (x+2y) (x-2y)​
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Pop ok you see so the answer here would be frantically elastic
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2 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}}

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5 times as many stakes
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