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stiv31 [10]
3 years ago
7

When ∆RST is translated 4 units down, what are the apparent coordinates of T’?

Mathematics
1 answer:
Klio2033 [76]3 years ago
4 0

Answer:

T' is at (-1,-8)

Step-by-step explanation:

When we translate down 4 units, we will subtract 4 from the y coordinate

T is at (-1,-4)

We need to subtract 4 from the y coordinate

T' is at (-1,-4-4)

T' is at (-1,-8)

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Which is the equation in the slope-intercept form of the line that contains points E & F ?
sesenic [268]

Answer:

y = 4x - 12

Step-by-step explanation:

The equation of a line in slope- intercept form is

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3 years ago
Read 2 more answers
Calculate the sum of the multiples of 4 from 0 to 1000
allochka39001 [22]

Answer:

sum is 125,500

sum in summation notation is \sum\limits_{n=0}^n a+nd= (2a+(n-1)d)n/2

Step-by-step explanation:

This problem can be solved using concept of arithmetic progression.

The sum of n term terms in arithmetic progression is given by

sum = (2a+(n-1)d)n/2

where

a is the first term

d is the common difference of arithmetic progression

_____________________________________________________

in the problem

series is multiple of 4 starting from 4 ending at 1000

so series will look like

series: 0,4,8,12,16..................1000

a is first term so

here a is 0

lets find d the common difference

common difference is given by nth term - (n-1)th term

lets take nth term as 8

so (n-1)th term = 4

Thus,

d = 8-4 = 4

d  can also be seen 4 intuitively as series is multiple of four.

_____________________________________________

let calculate value of n

we have last term as 1000

Nth term can be described

Nth term = 0+(n-1)d

1000 =   (n-1)4

=> 1000 = 4n -4

=> 1000 + 4= 4n

=> n = 1004/4 = 251

_____________________________________

now we have

n = 1000

a = 0

d = 4

so we can calculate sum of the series by using formula given above

sum = (2a+(n-1)d)n/2

       = (2*0 + (251-1)4)251/2

       = (250*4)251/2

     = 1000*251/2 = 500*251 = 125,500

Thus, sum is 125,500

sum in summation notation is \sum\limits_{n=0}^n a+nd= (2a+(n-1)d)n/2

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