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2 years ago

Write the standard form of the equation of the line passing through the point (-4,1) and perpendicular to the line

Mathematics

1 answer:

notka56 [123]2 years ago

5 0

Answer:

all work is shown/pictured

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2 years ago

Please help me with my math

WARRIOR [948]

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2 years ago

The length of a rectangle is 3 less than twice the width. Write a function rule for the perimeter in terms of w, the width

Nitella [24]

l = 2w-3

P = 2(l+w)

substitute in for l

P = 2(2w-3 +w)

combine like terms

P = 2(3w-3)

P = 6w-6

P(w) = 6(w-1)

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2 years ago

Read 2 more answers

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

2 years ago

Read 2 more answers

-6(y + 15) = -3y + 6 what is the value of y

Serga [27]

Let's begin by listing out the information given to us:

$-6\mleft(y+15\mright)=-3y+6$

We are to find the value of y

$undefined$

6 0

5 months ago