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Mathematics

2 answers:

r-ruslan [8.4K]3 years ago

6 0

Answer:

$y = - \frac{1}{2} x + 1$

Over [174]3 years ago

4 0

Answer:

y=-1/2x+1

Step-by-step explanation:

The slope of the line can be found using the rise over run. The rise is how far it goes up or down the y axis and the run is how far it goes left or right on the x-axis. Then put the rise over run and you have your slope.The rise is going down 1, so the value for rise is -1. The run is going 2 to the right, to the value for run is 2. Then rise over run, you end up with -1/2. To find the y-intercept, look at how many units the line is up from the x-axis on the y axis. The line is 1 unit above the x-axis on the y axis, so the y-intercept is 1. Puting the equation together, you get y=-1/2x+1.

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Arte-miy333 [17]

Answer:

A. too much

Step-by-step explanation:

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3 years ago

Perpendicular to y=1/2-1, x-intercept at4

AlexFokin [52]

Answer:

it would be Y= -2x+4

Step-by-step explanation:

To begin, we know that the b value is 4, because that is the x-intercept that is information given, so we can atomatically put that in.

Next we need to find the m value, which can be found by knowing the rules of a perpendicular eqution. to change it you must change the 1/2 and flip the equation and change the signs

soooooo 1/2 would change into 2/1 which is 2 and we would addd the negative sign so

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8 0

3 years ago

Read 2 more answers

What is the interquartile range of this data set 2, 5, 9, 11, 18, 30, 42, 48, 71, 73, 81

Shtirlitz [24]

Answer:

I believe the answer is 62.

Step-by-step explanation:

Hope my answer has helped you!

8 0

3 years ago

Using a linear approximation, estimate f(2.1), given that f(2) = 5 and f'(x) = √3x-1.

algol [13]

Answer:

$f\left( {2.1} \right) \approx 5.22360$ .

Step-by-step explanation:

The linear approximation is given by the equation

${f\left( x \right) \approx L\left( x \right) }={ f\left( a \right) + f^\prime\left( a \right)\left( {x - a} \right).}$

Linear approximation is a good way to approximate values of $f(x)$ as long as you stay close to the point $x= a$ , but the farther you get from $x=a$ , the worse your approximation.