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IrinaK [193]
3 years ago
7

If you want points for freee.

Mathematics
2 answers:
romanna [79]3 years ago
6 0

Answer:

WHA

Step-by-step explanation:

dalvyx [7]3 years ago
6 0

Answer:

Thank you vroo

Step-by-step explanation:

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$1.90

Step-by-step explanation:

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Tìm m để phương trình có 1 nghiệm duy nhất<br> x+2y-2z=2<br> 3x+7y-z=5<br> 2x+4y+mz=7
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Rewrite the system of equations in matrix form.

\begin{bmatrix}1&2&-2\\3&7&-1\\2&4&m\end{bmatrix} \mathbf x = \mathbf b

This system has a unique solution \mathbf x = \mathbf A^{-1}\mathbf b so long as the inverse of the coefficient matrix \mathbf A exists. This is the case if the determinant is not zero.

We have

\det(\mathbf A) = m+4

so the inverse, and hence a unique solution to the system of equations, exists as long as m ≠ -4.

4 0
3 years ago
Find the perimeter and area
LekaFEV [45]

Answer:

Perimeter = 156

Area = 1029

Step-by-step explanation:

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P = 156

Area = 49*42/2

A = 1029

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3 years ago
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What is 199 times 347
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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

3 0
3 years ago
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