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zavuch27 [327]
3 years ago
10

What is the x-intercept of 3x – 4y = 242

Mathematics
1 answer:
Andreas93 [3]3 years ago
5 0

Answer:

x=242/3

alternative form= 80 2/3, x=80.6

Step-by-step explanation:

3x-4y=242

substitute y = 0

3x-4x0=242

calculate the product

3x-0=242

remove 0

3x=242

divide both sides (by 3 in order to leave x by itself)

done

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(a) Use the fundamental theorem of algebra to determine the number of roots for 2x^2 + 4x + 7
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Answer:

Step-by-step explanation:

A 2nd order polynomial such as this one will have 2 roots; a 3rd order polynomial 3 roots, and so on.

The quadratic formula is one of the faster ways (in this situation, at least) in which to find the roots.  From 2x^2 + 4x + 7 we get a = 2, b = 4 and c = 7.

Then the discriminant is b^2 - 4ac, or, here, 4^2 - 4(2)(7), or -40.  Because the discriminant is negative, we know that the roots will be complex and unequal.

Using the quadratic formula:

       -4 ±√[-40]         -4 ± 2i√10

x  = ------------------ = ------------------

                4                       4

                                       -2 ± i√10

Thus, the roots are x = ------------------

                                               2

4 0
3 years ago
Please answer correctly !!!!!!!!! Will<br> Mark brainliest !!!!!!!!!!!!!
il63 [147K]

Answer:

x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}

Step-by-step explanation:

w=-5\left(x-8\right)\left(x+4\right)\\\mathrm{Expand\:}-5\left(x-8\right)\left(x+4\right):\quad -5x^2+20x+160\\w=-5x^2+20x+160\\Switch\:sides\\-5x^2+20x+160=w\\\mathrm{Subtract\:}w\mathrm{\:from\:both\:sides}\\-5x^2+20x+160-w=w-w\\Simplify\\-5x^2+20x+160-w=0\\Solve\:with\:the\:quadratic\:formula\\\mathrm{Quadratic\:Equation\:Formula:}\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=-5,\:b=20,\:c=160-w:\quad x_{1,\:2}=\frac{-20\pm \sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}\\x=\frac{-20+\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20+\sqrt{-20w+3600}}{10}\\x=\frac{-20-\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20-\sqrt{-20w+3600}}{10}\\The\:solutions\:to\:the\:quadratic\:equation\:are\\x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}

6 0
3 years ago
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