Simplify the absolute value 4

PLEASE HELP!!!!

Lunna [17]

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 UP
if "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 UP

x = $\frac{-b}{2a}$ = $\frac{-11}{2(1)}$ = $-\frac{11}{2}$

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 DOWN

x = $\frac{-b}{2a}$ = $\frac{-(-6)}{2(-1)}$ = $\frac{6}{-2}$ = -3

y = -(-3)² - 6(-3) - 8 = -9 + 18 - 8 = 1

vertex (-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 UP

x = $\frac{-b}{2a}$ = $\frac{-(-2)}{2(1)}$ = $\frac{2}{2}$ = 1

y = (1)² - 2(1) + 3 = 1 - 2 + 3 = 2

vertex (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 UP

x = $\frac{-b}{2a}$ = $\frac{-4}{2(1)}$ = -2

y = (-2)² + 4(-2) + 4 = 4 - 8 + 4 = 0

vertex (-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 UP

x = $\frac{-b}{2a}$ = $\frac{-21}{2(3)}$ = $-\frac{7}{2}$

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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4 0

4 years ago

Read 2 more answers

1 2 3 4 5 6 7 8 9 10 TIME REMAINING 54:32 Because the lid of a marker is wider than the marker itself, a set of markers can be p

tia_tia [17]

Answer:

10 Square Inches

Step-by-step explanation:

Trapezoid

Base =12 inches

Height =10 inches

Top side= 10 inches.

Area of a Trapezoid $=\frac{1}{2}(a+b)h, $where a and b are the lengths of the base and top respectively$$

$=\frac{1}{2}(10+12)*10\\=5*22=110 \:Square\:Inches$

Area of the Trapezoid=110 Square Inches

Rectangle

Base = 12 inches

Height =10 inches.

Area of a Rectangle=Base X Height

=12 X 10

=120 Square Inches

Difference in Area between the two packages

Difference=Area of Rectangle-Area of Trapezoid

=120-110

=10 Square Inches

5 0

3 years ago

Read 2 more answers

An engineer, in an attempt to make a quick measurement, walks 100 ft from the base of an overpass

Korvikt [17]

Answer:

The distance of the overpass above the ground is approximately 26.795 ft

Step-by-step explanation:

The parameters given are;

The distance from the overpass the engineer stands before determining the angle of elevation of the overpass from his standing point = 100 ft

The angle of elevation of the overpass as determined by the engineer from 100 ft = 15°

By trigonometric ratios, we have;

$Tan(\theta) = \dfrac{Opposite \, side \, to\ angle}{Adjacent\, side \, to\, angle}$

The opposite side to the 15° angle of elevation in the above case is the distance of the overpass above the ground

The opposite side to the 15° is the distance of the engineer from the base of the overpass

Therefore;

Tan(15°) the height of the overpass=

length

$Tan(15 ^{\circ}) = \dfrac{The \ distance \, of \, the \ overpass \ above \ ground}{100 \ ft}$

The distance of the overpass above the ground = 100 × tan (15°) ≈ 26.795 ft.

6 0

3 years ago

Can someone help me

nexus9112 [7]

Answer:

Its C i think

Step-by-step explanation:

3 0

3 years ago

The question is in photos

DedPeter [7]

Answer: 2535

Step-by-step explanation:

use order of operations (parenthesis, exponents, multiplication/division, addition/subtraction)

first ill handle the two innermost parenthesis

(4 + 3(2 - 8))

(4 + 3(-6))

(4 - 18)

-14

now lets move on to the next parenthesis

( 3 - 5^2(-14))

(3 - 25( - 14)

( 3 + 350)

353

now onto the last part of the problem

8^2 + 7 (353)

64 + 2471

2535 :))

3 0

3 years ago

Simplify the absolute value 4​

Simplify the absolute value 4