I need help with this math question on my study guide!Sadly, I do not remember how to do this and I would greatly appreciate any help you can offer.

Answers

Answer 1

Answer: Sadly is right. You must have practiced this stuff in middle school
until it was coming out of your ears. Fortunately, you have a chance
to relearn it now. You should do that ... it'll be important in any math
course you ever take.

'm' is the slope of the line on the graph.

The slope is

      (the change in 'y' between any two points on the line)
divided by
   (the change in 'x' between the same two points).

You can choose any two points on the line, and the slope is always the same.

To make it easy, look at the two points on this graph where the line crosses
the x-axis and the y-axis.

Going between these two points ...
-- the line goes up, from y=0 to y=4. The change in 'y' is 4 .
-- the line goes to the right, from x=-2 to x=0. The change in 'x' is 2 .

   'm' = the slope = (4)/(2) = 2 .

The "y-intercept" is the place where the line crosses the y-axis.
On this graph, that's the point where y=4 .

The equation of EVERY straight line on ANY graph is:

   Y = (the slope) times 'x' + (the y-intercept) .

So the equation of THIS line on THIS graph is

   Y = 2x + 4 .

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What is the approximate value of the function at x = -3?

Answers

The answer is that u have to see what = 3 equals three so find that out and u have ur answer

The daily high temperatures, in degrees Celsius, for two weeks were recorded. The first week had a mean high temperature of 8 degrees Celsius and a mean absolute deviation (MAD) of 1 degree Celsius. The second week had a mean high temperature of 4.86 degrees Celsius and a mean absolute deviation (MAD) of 1.91 degrees Celsius. Which week had greater variability in high temperature? Explain your reasoning.

Answers

The required week that had greater variability in high temperature is week 2.

Given that,
The first week had a mean high temperature of 8 degrees Celsius and a mean absolute deviation (MAD) of 1 degree Celsius.
The second week had a mean high temperature of 4.86 degrees Celsius and a mean absolute deviation (MAD) of 1.91 degrees Celsius.

What is Statistic?

Statistics is the study of mathematics that deals with relations between comprehensive data.

Here,
The system of observation that has a higher mean absolute deviation, will have the chance of greater variability in high temperature. So,
mean absolute deviation of week 1 = 1
mean absolute deviation of week 2 = 1.91
MAD of week 2 > MAD of week 1

Thus, the required week that had greater variability in high temperature is week 2.

Learn more about Statistics here:
brainly.com/question/23091366

#SPJ2

Answer:

mean 4.86

MAD = 1.91

Step-by-step explanation:

Will 3/5 x 2/7 be less than, greater than or equal to 3/5

Answers

3/5 times 2/7=6/35
compare 3/5 to 6/35

5 times 7=35

3/5 times 7/7=21/35

21/35 ? 6/35
21 ? 6
21> 6

an easy way is if we do

3/5=x, a random number
rememberig that taking a fraction of something makes it smaller

x times fraction is smaller than x
so
3/5 times 2/7 is less than 3/5

3/5×2/7=6/35
3/5=3/5×7/7=15/35
3/5 ×2/7 is less than 3/5

Ten people are entered in a race. If there are no ties, in how many ways can the first three places come out?A. 1440

B. 720

C. 30

D. 132

Answers

B) 720 different possibilities

Answer: (720) is the right answer

Step-by-step explanation:

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

Dimtry bought a pair of pants at the discounted price of $30. The original price of the pants was $40. What was the percent of the discount? PLEASE EXPLAIN TO ME HOW YOU GOT YOUR ANSWER

Answers

Answer:

25%

Step-by-step explanation:

Say we start from the beginning with $40. If the pants are discounted but we don't know the percent discount, then we need a variable, x:

40 - x% * 40 = 30

The total is 40. When we take a discount off of 40, we are taking a percent of 40 and subtracting that from 40. From the problem, this should be 30. Thus, the equation above.

Add x%*40 to both sides and subtract 30 from both sides:

x%*40 = 10

Divide by 40:

x% = 1/4 = 0.25

Multiply each side by 100:

x = 25

The answer is 25%.

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