In a function you cannot have to of the same x values right?

Answers

Answer 1

Answer: Yes, in a function you cannot have two different images for the same x.

Because if an x has more than one image, you couldn't tell what is the value of the image given that x.

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9.8 < 20 – 9.8t < 20

In which quadrant is (1/2, -1.8)?Quadrant II
Quadrant IV
Quadrant III
Quadrant I

Answers

Answer:

Quadrant lV

Step-by-step explanation:

Quadrant l = (+,+)

Quadrant ll = (-,+)

Quadrant lll = (-,-)

Quadrant lV = (+,-)

This is how you can always remember, hope this helps!

Which of the following tables has a slope of zero?

Answers

Answer:

Step-by-step explanation:

The Surf Shop gives surfinglessons to 72 people. If the
shop gives lessons to groups
of 5 people, how many total
groups will there be?

Answers

Answer:

Step-by-step explanation:

Just divide 72 by 5 because of there being 72 total peoples and them being DIVIDED into groups of 5.

72/5=14.4

Though you cant have a .4 group of people, so there would be 15 groups.

The graph of which function has an axis of symmetry at x =-1/4 ?f(x) = 2x2 + x – 1

f(x) = 2x2 – x + 1

f(x) = x2 + 2x – 1

f(x) = x2 – 2x + 1

Answers

we know that

The equation of the vertical parabola in vertex form is equal to

$y=a(x-h)^(2)+k$

where

(h,k) is the vertex

The axis of symmetry is equal to the x-coordinate of the vertex

$x=h$ ------> axis of symmetry of a vertical parabola

we will determine in each case the axis of symmetry to determine the solution

case A) $f(x)=2x^(2)+x-1$

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)+1=2x^(2)+x$

Factor the leading coefficient

$f(x)+1=2(x^(2)+0.5x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$f(x)+1+0.125=2(x^(2)+0.5x+0.0625)$

$f(x)+1.125=2(x^(2)+0.5x+0.0625)$

Rewrite as perfect squares

$f(x)+1.125=2(x+0.25)^(2)$

$f(x)=2(x+0.25)^(2)-1.125$

the vertex is the point $(-0.25,-1.125)$

the axis of symmetry is

$x=-0.25=-(1)/(4)$

therefore

the function $f(x)=2x^(2)+x-1$ has an axis of symmetry at $x=-(1)/(4)$

case B) $f(x)=2x^(2)-x+1$

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)-1=2x^(2)-x$

Factor the leading coefficient

$f(x)-1=2(x^(2)-0.5x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$f(x)-1+0.125=2(x^(2)-0.5x+0.0625)$

$f(x)-0.875=2(x^(2)-0.5x+0.0625)$

Rewrite as perfect squares

$f(x)-0.875=2(x-0.25)^(2)$

$f(x)=2(x-0.25)^(2)+0.875$

the vertex is the point $(0.25,0.875)$

the axis of symmetry is

$x=0.25=(1)/(4)$

therefore

the function $f(x)=2x^(2)-x+1$ does not have a symmetry axis in $x=-(1)/(4)$

case C) $f(x)=x^(2)+2x-1$

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)+1=x^(2)+2x$

Complete the square. Remember to balance the equation by adding the same constants to each side

$f(x)+1+1=x^(2)+2x+1$

$f(x)+2=x^(2)+2x+1$

Rewrite as perfect squares

$f(x)+2=(x+1)^(2)$

$f(x)=(x+1)^(2)-2$

the vertex is the point $(-1,-2)$

the axis of symmetry is

$x=-1$

therefore

the function $f(x)=x^(2)+2x-1$ does not have a symmetry axis in $x=-(1)/(4)$

case D) $f(x)=x^(2)-2x+1$

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)-1=x^(2)-2x$

Complete the square. Remember to balance the equation by adding the same constants to each side

$f(x)-1+1=x^(2)-2x+1$

$f(x)=x^(2)-2x+1$

Rewrite as perfect squares

$f(x)=(x-1)^(2)$

the vertex is the point $(1,0)$

the axis of symmetry is

$x=1$

therefore

the function $f(x)=x^(2)-2x+1$ does not have a symmetry axis in $x=-(1)/(4)$

the answer is

$f(x)=2x^(2)+x-1$

axis of symmetry is the x value of the vertex

for
y=ax^2+bx+c
x value of vertex=-b/2a

first one
-1/2(2)=-1/4
wow, that is right

answer is first one
f(x)=2x^2+x-1

How to solve for y if the equation is y=5/2x + 5 if x is zeroplease show step by step how to solve

Answers

$y=(5)/(2)\cdot0+5\ny=0+5\ny=5$

Q, y=5/2x+x if x is zero

An, ( y=5/2×0+5 ) as you know if i multiply number to zero you get zero so "x" change to the zero you get
( y=0+5 )= ( y=5 ) that is your answer

The price of a chromebook is $230. The sales tax is 12%. What is the amount of the sales tax for the chromebook?$57.20

$9.60

$23.20

$27.60

Answers

Answer:

27.60

Step-by-step explanation:

take your $230 and takes 12% and do 12% of 230 which will get you 27.60

27.60 you have to multiply 230 x 12 and there is your answer

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