Mathematics, 02.07.2019 11:30 nailah742

Jeremy and arnold are working on a project for math class in which they are to identify the quadratic equation that represents a rain gauge that sits off the ground. graph each of the equations and then determine which one could represent the position of the rain catcher that sits above the ground. the x-axis represents the ground.

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Answer from: Heyitsbrandi

NOTES:

Given a quadratic function in standard format (y = ax² + bx + c), the direction of the parabola is as follows:

if "a" is positive, then opens UPif "a" is negative, then opens DOWN

Given a quadratic function in standard format (y = ax² + bx + c), the vertex can be found as follows:

the Axis Of Symmetry (x-value) is: x = $\frac{-b}{2a}$ y-value is found by plugging in the AOS for "x" in the equation

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1) y = x² + 11x + 24

a = +1 so the parabola opens UPx = $\frac{-b}{2a}$ = $\frac{-11}{2(1)}$ = $-\frac{11}{2}$ y = $(-\frac{11}{2})^{2}$ + 11 $(-\frac{11}{2} )$ + 24 = $-\frac{25}{4}$ vertex $(-\frac{11}{2}$ , $-\frac{25}{4})$ is in Quadrant 3 and is below the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-3, 0) and (-8, 0), and y-intercept (0, 24)

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2) y = -x² - 6x - 8

a = -1 so the parabola opens DOWNx = $\frac{-b}{2a}$ = $\frac{-(-6)}{2(-1)}$ = $\frac{6}{-2}$ = -3y = -(-3)² - 6(-3) - 8 = -9 + 18 - 8 = 1vertex (-3, 1) is in Quadrant 2 and is above the x-axis

This could NOT be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-2, 0) and (-4, 0), and y-intercept (0, -8)

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3) y = x² - 2x + 3

a = 1 so the parabola opens UPx = $\frac{-b}{2a}$ = $\frac{-(-2)}{2(1)}$ = $\frac{2}{2}$ = 1y = (1)² - 2(1) + 3 = 1 - 2 + 3 = 2vertex (1, 2) is in Quadrant 1 and is above the x-axis

This could NOT be the graph of the rain gauge.

The graph should contain the vertex, y-intercept (0, 3), and its mirror image (2, 3). There are no x-intercepts

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4) y = x² + 4x + 4

a = 1 so the parabola opens UPx = $\frac{-b}{2a}$ = $\frac{-4}{2(1)}$ = -2y = (-2)² + 4(-2) + 4 = 4 - 8 + 4 = 0vertex (-2, 0) is in Quadrant 2 and is on the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, y-intercept (0, 4), and its mirror image (-4, 4). The x-intercept is the vertex.

Compared to the other four graphs, this is most likely the equation for the rain gauge!

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5) y = 3x² + 21x + 30

a = +3 so the parabola opens UPx = $\frac{-b}{2a}$ = $\frac{-21}{2(3)}$ = $-\frac{7}{2}$ y = 3 $(-\frac{7}{2})^{2}$ + 21 $(-\frac{7}{2} )$ + 30 = $-\frac{27}{4}$ vertex $(-\frac{7}{2}$ , $-\frac{27}{4})$ is in Quadrant 3 and is below the x-axis

This COULD be the graph of the rain gauge.

The graph should contain the vertex, x-intercepts (-2, 0) and (-5, 0), and y-intercept (0, 30)

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Answer from: Quest

e^-kt = n/n0 = 0.79

-kt = ln 0.79

t = - ln 0.79/0.0001

= 2357 years

step-by-step explanation:

Answer from: Quest

Hello!

the answer is:

the lateral area is equal to $352m^{2}$

$lateralsidearea=352m^{2}$

why?

to find the lateral area of the square pyramid, first we need to find the perimeter of the base, and then, find the lateral area using the following formulas:

$baseperimeter=4s$

$lateralsidearea=\frac{1}{2}baseperimeter*slantheight$

where,

s, is the base length

slant height, is the height of a the side of the pyramid

we are given:

$s=8m\\slantheight=22m$

now, calculating the perimeter of the base, we have:

$baseperimeter=4s=4*8m=32m$

then, calculating the lateral area:

$lateralsidearea=\frac{1}{2}baseperimeter*slantheight$

$lateralsidearea=\frac{1}{2}32*22m=352m^{2}$

hence, we have that the lateral area is equal to $352m^{2}$

have a nice day!

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