All categories

3 years ago

Determine whether the equation is true or false. sum x=2 ^ 8 (2n-3)= sum n=3 ^ 9 (2m-5)

Mathematics

1 answer:

Sidana [21]3 years ago

6 0

Answer: true

Step-by-step explanation:

You might be interested in

Figures ABCD and PQRS are parallelogram find x and y (lengths are in metres) Part i and ii.

STatiana [176]

Step-by-step explanation:

For 1st question

In a parm ABCD

5x - 4 = 16 ( being opposite sides of parallelogram)

5x = 16 + 4

5x = 20

x = 20 / 5

x = 4

3y + 1 = 22 (being opposite sides of parallelogram)

3y = 22 -1

3y = 21

y = 21 / 3

y = 7

For second question

In parm PQRS

Diagonals bisect each other so

x + 4 = 12

x = 12 - 4

x = 8

x + y = 32

8 + y = 32

y = 32 - 8

y = 24

Hope it will help :)❤

8 0

3 years ago

7 3/9 - 2 4/9 = 11 3/4 - 5 1/4 =

jeka57 [31]

1.)

$7 \frac{3}{9} - 2 \frac{4}{9} \\ = \frac{66}{9} - \frac{22}{9} \\ = \frac{44}{9} \\ = 4 \frac{8}{9}$

2.)

$11 \frac{3}{4} - 5 \frac{1}{4} \\ = \frac{47}{4} - \frac{21}{4} \\ = \frac{26}{4} \\ = \frac{13}{2} \\ = 6 \frac{1}{2}$

Hope it helps and is useful :)

6 0

3 years ago

Drag each tile to the table to multiply (6x-y)(2x-y+2)

77julia77 [94]

The other answer is 12x2 – 8xy + 12x + y2 – 2y

7 0

2 years ago

I'm really bad at math so I need some help explaining.

sergiy2304 [10]

Please let me know if there’s something you don’t understand. Hope this helps!

3 0

1 year ago

Hey guys im new here please solve this for me with steps! ill mark as the best answer

Vinil7 [7]

Answer:

The factors of $2(x+y)^2-9(x+y)-5$ is ((x+y)-5)(2x+2y+1)

Step-by-step explanation:

Given polynomial

=> $2(x+y)^2-9(x+y)-5$

To Find:

The factors of the polynomial =?

Solution:

Lets assume k = (x+y)

Then $2(x+y)^2-9(x+y)-5$ can be written as $2k^2-9k-5$

Now by using quadratic formula

k = $\frac{-b\pm\sqrt{(b^2-4ac}}{2a}$

where

a= 2

b= -9

c= -5

Substituting the values, we get

k = $\frac{-b\pm\sqrt{(b^2-4ac)}}{2a}$

k = $\frac{-(-9) \pm \sqrt{((-9)^2-4(2)(-5)}}{2(2))}$

k = $\frac{-(-9) \pm \sqrt{(81+40)}}{4}$

k = $\frac{-(-9) \pm \sqrt{(121)}}{4}$

k = $\frac{-(-9) \pm 11}}{4}$

k= $\frac{ 9 \pm 11}{4}$

k = $\frac{20}{4}$ k = $\frac{-2}{4}$

$k_1 =5$ $k_2 = -\frac{1}{2}$

$2k^2-9k-5= 2(k-5)(k+\frac{1}{2})$

Solving the RHS we get

$\frac{2}{2}(k-5)(2k+1)$

$(k-5)(2k+1)$

Substituting k = x+y

$((x+y)-5)(2(x+y+1)$

$((x+y)-5)(2x+2y+1)$

5 0

2 years ago