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2 years ago

5

Give the slope between the points (-5,-3) (-1,-8)

1 answer:

Dima020 [189]2 years ago

4 0

Slope is difference in y over difference in x:

m = ( -8 - -3) / (-1 - -5) = (-5)/(4) = -5/4

The slope is -5/4.

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2. Find the gross monthly income from this job and calculate your monthly take-home pay.

Step2247 [10]

Answer:

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4 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

Determine the value of x.<br><br><br><br> 20°<br> 30°<br> 45°<br> 60°

mars1129 [50]

30 is the value of x given of above quertion

4 0

2 years ago

Read 2 more answers

Compute the price of a 3.8 percent coupon bond with 18 years left to maturity and a market interest rate of 7 percent

zzz [600]

The photo below may help you

5 0

3 years ago

3x^3-2x^2-147x+98=(ax-c)(bx+d)(bx-d) where a,b,c and d positive integers. Work out the value of a,b,c and d

Alexxx [7]

Expand the right side:

$(ax-c)(bx+d)(bx-d)=(ax-c)(b^2x^2-d^2)$

$=ab^2x^3-b^2cx^2-ad^2x+cd^2$

Notice that 98 = 2 * 7 * 7, so we have $c=2$ and $d=7$ .

Then

$-ad^2=-147\implies a=\dfrac{147}{49}=3$

and

$ab^2=3\implies b^2=1\implies b=1$

6 0

2 years ago