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NNADVOKAT [17]
3 years ago
8

Find the missing measurements of the right triangle. Round your answers to the

Mathematics
1 answer:
nadya68 [22]3 years ago
4 0

Answer:

m∠B = 63°

AB = 26.4

AC = 23.6

Step-by-step explanation:

m∠B = 90 - 27 = 63°

Sin 27 = 12/AB

AB (sin 27) = 12

AB = 12/sin 27 = 26.4

Tan 27 = 12/AC

AC (tan 27) = 12

AC = 12/tan 27 = 23.6

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Find the x-intercept of the parabola with vertex (1,-13) and y-intercept (0,-11).
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1. find equation

general solutions are:
y = a x^{2} +bx +c \\ \\ \frac{dy}{dx} = 2ax +b

for P(0,-11)

-11 = 0a + 0b +c ⇒c = -11

for p(1,-13)
-13 = 1a + 1b -11 \\ a+b = -2

and

0 = 2a + b \\ b = -2a

solve for a and b:
a - 2a = -2 \\ a = 2 \\ b = -4

the total equation is now:

y = 2 x^{2} -4x -11

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This is the solution

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2 years ago
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Let p(x) = x^2 + 8x + 12 and q(x) = x^3 + 5x^2 - 6x. Define r(x) = p(x) \cdot q(x).
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The roots of the entire <em>polynomic</em> expression, that is, the product of p(x) = x^2 + 8x + 12 and q(x) = x^3 + 5x^2 - 6x, are <em>x₁ =</em> 0, <em>x₂ =</em> -2, <em>x₃ =</em> -3 and <em>x₄ =</em> -6.

<h3>How to solve a product of two polynomials </h3>

A value of <em>x</em> is said to be a root of the polynomial if and only if <em>r(x) =</em> 0. Let be <em>r(x) = p(x) · q(x)</em>, then we need to find the roots both for <em>p(x)</em> and <em>q(x)</em> by factoring each polynomial, the factoring is based on algebraic properties:

<em>r(x) =</em> (x + 6) · (x + 2) · x · (x² + 5 · x - 6)

<em>r(x) =</em> (x + 6) · (x + 2) · x · (x + 3) · (x + 2)

r(x) = x · (x + 2)² · (x + 3) · (x + 6)

By direct inspection, we conclude that the roots of the entire <em>polynomic</em> expression are <em>x₁ =</em> 0, <em>x₂ =</em> -2, <em>x₃ =</em> -3 and <em>x₄ =</em> -6. \blacksquare

To learn more on polynomials, we kindly invite to check this verified question: brainly.com/question/11536910

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Theorems! Please help. Problem in picture.
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I would review this back again later.
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