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2 years ago

The vertices of a rectangle are R(-5,-5), S(-1,-5), T(-1,1) and U(-5,1). After translation R' is the point (-3,-8). What are the

translation and rule coordinates of U? A. (x+2,y+3); (-3,4) B. (x+2,y-3); (-3,-2) C. (x-2,y-3); (-7,-2) D. (x-2,y+3); (-7,4)

Mathematics

1 answer:

3241004551 [841]2 years ago

7 0

$R(-5;-5)\to R'(-3;-8)\\\\x:-3-(-5)=-3+5=2\\y:-8-(-5)=-8+5=-3\\\\Your\ answer:\boxed{B.\ (x+2;\ y-3)\to U'(-5+2;\ 1-3)\to U'(-3;-2)}$

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The function h(t) = −16t2 + 18t models the height, in feet, reached by a leopard t seconds after it jumps. What is the approxima

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9514 1404 393

Answer:

5 1/16 ft

Step-by-step explanation:

h(t) = -16t(t -18/16) . . . . put in intercept form

The function describes a parabola that opens downward. It has zeros at t=0 and t=9/8. The maximum height will be found at the vertex of the parabola, halfway between these zeros.

f(9/16) = (-16)(9/16)² +18(9/16) = 81/16 = 5 1/16 . . . . feet

The approximate maximum height of the leopard is 5 1/16 feet.

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3 years ago

Pleasantburg has a population growth model of P(t)=at2+bt+P0 where P0 is the initial population. Suppose that the future populat

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Answer:

The population will reach 34,200 in February of 2146.

Step-by-step explanation:

Population in t years after 2012 is given by:

$P(t) = 0.8t^{2} + 6t + 19000$

In what month and year will the population reach 34,200?

We have to find t for which P(t) = 34200. So

$P(t) = 0.8t^{2} + 6t + 19000$

$0.8t^{2} + 6t + 19000 = 34200$

$0.8t^{2} + 6t - 15200 = 0$

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

$0.8t^{2} + 6t - 15200 = 0$

So $a = 0.8, b = 6, c = -15200$

Then

$\bigtriangleup = 6^{2} - 4*0.8*(-15100) = 48356$

$t_{1} = \frac{-6 + \sqrt{48356}}{2*0.8} = 134.14$

$t_{2} = \frac{-6 - \sqrt{48356}}{2*0.8} = -141.64$

We only take the positive value.

134 years after 2012.

.14 of an year is 0.14*365 = 51.1. The 51st day of a year happens in February.

So the population will reach 34,200 in February of 2146.

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2 years ago

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3 years ago