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SpyIntel [72]
3 years ago
5

Can someone please help☹️❤️

Mathematics
1 answer:
dalvyx [7]3 years ago
8 0

Answer:

Slope = 2      y - intercept = ( 0 , -2 )      equation:  y = 2x - 2

Step-by-step explanation:

We can begin solving by first understanding what slope is.

Slope is the rate of change in a line.

The formula for slope is:

(For any two points on a line)

m =  y₁ - y₂  /  x₁ - x₂

**m is the variable for slope

Two points on the line are: ( 6, 10 ) and ( 2, 2 )

Here is where we'll plug in the formula:

m = 10 - 2  /  6 - 2

m = 8  /  4

m = 2

Now that we have our slope, we only have to look at the graph to find the y-intercept.

The y-intercept is the point on the graph crosses the y-axis (this effectively means that the x value will always be 0)

When we look at the graph, we can see that the y-intercept is: ( 0, -2 )

All that's left is to right the equation in slope-intercept form.

The equation is arranged just as it says in the name:

y = mx + b

(m = slope   b = y -intercept)

When we plug in our variables, the equation becomes:

y = 2x - 2

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Tatiana [17]

If a $24 dvd is marked down by 30% then the cost of the dvd would be 70% of the original cost. 100% - 30% = 70%

70% of 24 can be expressed as 0.7 x 24.

The answer is $16.80.

Hope this helps :)

4 0
3 years ago
43-44 please and thanks
KatRina [158]
The easiest way to do the fractions is is you turn them to decimals.

43. >
44. >

You think it might be the othe way around, but it’s a negative. So if it’s -36 and -35, the -35 will be bigger. Think about the number line. -36 comes before -35.
4 0
3 years ago
The width of each of five continuous classes in a frequency distribution is 5 and the lower class-limit of the lowest class is 1
Helga [31]

Let x and y be the upper and lower class limit of frequency distribution.

Given, width of the class = 5

⇒ x-y= 5 …

Also, given lower class (y) = 10 On putting y = 10, we get

x – 10= 5 ⇒ x = 15 So, the upper class limit of the lowest class is 15

Hence, the upper class limit of the highest class

=(Number of continuous classes x Class width + Lower class limit of the lowest class)

= 5 x 5+10 = 25+10=35

Hence,’the upper class limit of the highest class is 35.

<em><u>Alternate Method</u></em><em><u>.</u></em>

After finding the upper class limit of the lowest class, the five continuous classes in a frequency distribution with width 5 are 10-15,15-20, 20-25, 25-30 and 30-35.

Thus, the highest class is 30-35..

<h3>Hence, the upper limit of this class is 35.</h3>

<h2>_____________________</h2>

4 0
3 years ago
Please I need help with the rest of my question.
eduard

Answer:

x-intercept (4,0)

y-intercept (0,10)

Step-by-step explanation:

3 0
3 years ago
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What is the solution to the linear equation? 2/5 + p = 4/5 + 3/5p ​
masha68 [24]

Answer:

p=1

Step-by-step explanation:

2/5 + p = 4/5 + 3/5p

Multiply each side by 5 to clear the fractions

5(2/5 + p) = 5(4/5 + 3/5p)

2 +5p = 4 + 3p

Subtract 3p from each side

2+5p-3p = 4+3p-3p

2+2p = 4

Subtract 2 from each side

2+2p -2 = 4-2

2p =2

Divide by 2

2p/2 = 2/2

p=1

4 0
3 years ago
Read 2 more answers
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