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3 years ago

HELP PLS ILL MARK BRAINLIST

Mathematics

1 answer:

SVETLANKA909090 [29]3 years ago

3 0

Answer:

i think its to late to answer... sorry i didnt see it sooner.

Step-by-step explanation:

ill take that brainliest

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If B is the midpoint of AC and AC=8x-20 find BC

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We know that AC=8x-20 and B is the mid-point of AC

⇒ AB = BC = $\frac{AC}{2}$

⇒ BC = $\frac{8x-20}{2}$

⇒ BC = 4x-10

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Use figure ABCD draw a line segment from point B to D. Name and classify the triangles formed

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The two triangles formed will be ADB and BCD

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What is 1 plus 6 mines 2 plus 5

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The value of y varies directly with x, and y = 48 when x = 16. Find y when x = 176.

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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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3 years ago

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