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3 years ago

Find g(9) if g(x) =2 (x+3)-4

Mathematics

1 answer:

ikadub [295]3 years ago

6 0

Answer:

Step-by-step explanation:

g(9)=2 since g(x)=2

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What is necessary to determine the z-score for a raw score in a particular set of scores?.

Xelga [282]

Answer:

See below

Step-by-step explanation:

The mean score and the standard deviation are needed

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2 years ago

Please help pls pls help pls belp

Delvig [45]

Answer:

After 1 second, the ball will reach a maximum height of 16 feet

Step-by-step explanation:

The height of the ball after t seconds: h(t) = -16t^2 + 32t

The graph of this quadratic function is parabola which opens downwards. The vertex of a quadratic equation is the maximum or minimum point on the equation's parabola

t = -b/2a = -(32)/2(-16) = -32/-32 = 1 second

then

h(t) = -16(1)^2 + 32(1) = -16 + 32 = 16

After 1 second, the ball will reach a maximum height of 16 feet

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2 years ago

Simon bought 3.5 pounds of Beef on sale for $1.25 per pound at HEB. If Simon gave the cashier $10.00, how much change would he r

Elenna [48]

Answer:

5.625

Step-by-step explanation:

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3 years ago

Paige was playing a trivia game where you gained points for correct answers and lost points for incorrect answers. At the start

victus00 [196]

Givens

Start = S = - 700

# Correct = C = 8

Value Correct = VC = 400

# Incorrect =I = 9

Value Incorrect =VI = - 600

Equation

Score = S + C*VC + I*VI Substitute

Solve

Score = -700 + 8*400 + 9(-600)

Score = -700 + 3200 - 5400

Score = - 700 - 2200

Score = - 2900 <<<<< Answer

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3 years ago

find the coordinates of the point P on the parabola y=1-x^2 with domain 0≤x≤1 that minimize the area of the triangle enclosed by

coldgirl [10]

Let point P be with coordinates $(x_0,y_0).$ Find the equation of the tangent line.

1. If $y=1-x^2,$ then $y'=-2x.$

2. The equation of the tangent line at point P is

$y-y_0=-2x_0(x-x_0).$

Find x-intercept and y-intercept of this line:

when x=0, then $y=y_0+2x_0^2;$
when y=0, then $x=\dfrac{y_0}{2x_0}+x_0=\dfrac{y_0+2x_0^2}{2x_0}.$

The area of the triangle enclosed by the tangent line at P, the x-axis, and y-axis is

$A=\dfrac{1}{2}\cdot (2x_0^2+y_0)\cdot \left(\dfrac{y_0+2x_0^2}{2x_0}\right)=\dfrac{(y_0+2x_0^2)^2}{4x_0}.$

Since point P is on the parabola, then $y_0=1-x_0^2$ and

$A=\dfrac{(1-x_0^2+2x_0^2)^2}{4x_0}=\dfrac{(1+x_0^2)^2}{4x_0}.$

Find the derivative A':

$A'=\dfrac{2(1+x_0^2)\cdot 2x_0\cdot 4x_0-4(1+x_0^2)^2}{16x_0^2}=\dfrac{12x_0^4+8x_0^2-4}{16x_0^2}.$

Equate this derivative to 0, then

$12x_0^4+8x_0^2-4=0,\\ \\3x_0^4+2x_0^2-1=0,\\ \\D=2^2-4\cdot 3\cdot (-1)=16,\ \sqrt{D}=4,\\ \\x_0^2_{1,2}=\dfrac{-2\pm4}{6}=-1,\dfrac{1}{3},\\ \\x_0^2=\dfrac{1}{3}\Rightarrow x_0_{1,2}=\pm\dfrac{1}{\sqrt{3}}.$

And

$y_0=1-\left(\pm\dfrac{1}{\sqrt{3}}\right)^2=\dfrac{2}{3}.$

Answer: two points: $P_1\left(-\dfrac{1}{\sqrt{3}},\dfrac{2}{3}\right), P_2\left(\dfrac{1}{\sqrt{3}},\dfrac{2}{3}\right).$

6 0

3 years ago

Find g(9) if g(x) =2 (x+3)-4​

Find g(9) if g(x) =2 (x+3)-4