Suppose you have an isosceles triangle, and each of the equal sides has a length of 1 foot. suppose the angle formed by those tw

o sides is 45. then the area of the triangle is

Mathematics

1 answer:

notsponge [240]3 years ago

8 0

$\left \{ {{y=2} \atop {x=2}} \right.$

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Find the sum of the squares of the roots of 3x^2 + 4x + 12 = 0

Contact [7]

Answer:

-56/9

Step-by-step explanation:

Let p and q be the roots of the equation.

So, here's how we're gonna work it out:

${p}^{2} + {q}^{2} = \\ {p}^{2} + {q}^{2} +2 pq - 2pq = \\ ( {p}^{2} +2pq + {q}^{2}) - 2pq = \\ {(p + q)}^{2} - 2pq$

Vieta's formulas give us the sum and the product of the roots of a quadratic equation. Using them we obtain:

p + q = -b / a = - 4 / 3
p × q = c / a = 12 / 3 = 4

${(p + q)}^{2} - 2pq = \\ {( - \frac{ 4}{3} )}^{2} - 2 \times 4 = \frac{16}{9 } - 8 = \\ \frac{16 - 72}{9} = - \frac{56}{9}$

4 0

3 years ago

Please help i’m so stuck i can’t do calc for the life of me

KATRIN_1 [288]

Answer:

$-3x^2+29x-150+\frac{946x^2-341x-756}{x^3+6x^2-3x-5}$ is the answer. See steps below.

Step-by-step explanation:

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}\\\\-3x^5+11x^4+33x^3-26x^2-36x-6\\\\\mathrm{and\:the\:divisor\:}x^3+6x^2-3x-5:\\\\\mathrm{Quotient}=\frac{-3x^5}{x^3}=-3x^2\\\\\mathrm{Multiply\:}x^3+6x^2-3x-5\mathrm{\:by\:}-3x^2\:\:\rightarrow\:\:-3x^5-18x^4+9x^3+15x^2\\\\\mathrm{Subtract\:}-3x^5-18x^4+9x^3+15x^2\mathrm{\:from\:}-3x^5+11x^4+33x^3-26x^2-36x-6\mathrm{\:to\:get\:new\:remainder}.\\\\\mathrm{Remainder}=29x^4+24x^3-41x^2-36x-6$