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Anton [14]
3 years ago
12

Answer please cuz I'm very confused

Mathematics
1 answer:
Anestetic [448]3 years ago
6 0

5x + 3y = - 53

the equation of a line in ' slope- intercept form ' is y = mx + c

where m is the slope and c the y-intercept

rearrange 3x - 5y = - 15 into this form to obtain m → (subtract 3x from both sides)

- 5y = - 3x - 15 → divide all terms by - 5 )

y = \frac{3}{5} x + 3 → in slope-intercept form with m = \frac{3}{5}

given a line with slope m then the slope m₁ of a line perpendicular to it is

m₁ = - \frac{1}{m} = - 1 ÷ \frac{3}{5} = - \frac{5}{3}

partial equation is y = - \frac{5}{3} x + c

to find c substitute ( - 7, - 6) into the partial equation

- 6 = \frac{35}{3} + c ⇒ c = - 6 - \frac{35}{3} = - \frac{53}{3}

y = - \frac{5}{3} x - \frac{53}{3} → in slope intercept form

multiply all terms by 3

3y = - 5x - 53 → ( add 5x to both sides )

5x + 3y = - 53 → in standard form





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1+-w2+9w and I need help cuz I’m on 76 and I’m sooo close help
Gnesinka [82]

\huge \boxed{\mathfrak{Question} \downarrow}

  • Simplify :- 1 + - w² + 9w.

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

\large \sf1 + - w ^ { 2 } + 9 w

Quadratic polynomial can be factored using the transformation \sf \: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where \sf x_{1} and x_{2} are the solutions of the quadratic equation \sf \: ax^{2}+bx+c=0.

\large \sf-w^{2}+9w+1=0

All equations of the form \sf\:ax^{2}+bx+c=0 can be solved using the quadratic formula: \sf\frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

\large \sf \: w=\frac{-9±\sqrt{9^{2}-4\left(-1\right)}}{2\left(-1\right)}  \\

Square 9.

\large \sf \: w=\frac{-9±\sqrt{81-4\left(-1\right)}}{2\left(-1\right)}  \\

Multiply -4 times -1.

\large \sf \: w=\frac{-9±\sqrt{81+4}}{2\left(-1\right)}  \\

Add 81 to 4.

\large \sf \: w=\frac{-9±\sqrt{85}}{2\left(-1\right)}  \\

Multiply 2 times -1.

\large \sf \: w=\frac{-9±\sqrt{85}}{-2}  \\

Now solve the equation \sf\:w=\frac{-9±\sqrt{85}}{-2} when ± is plus. Add -9 to \sf\sqrt{85}.

\large \sf \: w=\frac{\sqrt{85}-9}{-2}  \\

Divide -9+ \sf\sqrt{85} by -2.

\large \boxed{ \sf \: w=\frac{9-\sqrt{85}}{2}} \\

Now solve the equation \sf\:w=\frac{-9±\sqrt{85}}{-2} when ± is minus. Subtract \sf\sqrt{85} from -9.

\large \sf \: w=\frac{-\sqrt{85}-9}{-2}  \\

Divide \sf-9-\sqrt{85} by -2.

\large \boxed{ \sf \: w=\frac{\sqrt{85}+9}{2}}  \\

Factor the original expression using \sf\:ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \sf\frac{9-\sqrt{85}}{2}for \sf\:x_{1} and \sf\frac{9+\sqrt{85}}{2} for \sf\:x_{2}.

\large \boxed{ \boxed {\mathfrak{-w^{2}+9w+1=-\left(w-\frac{9-\sqrt{85}}{2}\right)\left(w-\frac{\sqrt{85}+9}{2}\right) }}}

<h3>NOTE :-</h3>

Well, in the picture you inserted it says that it's 8th grade mathematics. So, I'm not sure if you have learned simplification with the help of biquadratic formula. So, if you want the answer simplified only according to like terms then your answer will be ⇨

\large \sf \: 1 + -  w {}^{2}  + 9w \\  =\large  \boxed{\bf \: 1 -  {w}^{2}   + 9w}

This cannot be further simplified as there are no more like terms (you can use the biquadratic formula if you've learned it.)

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