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3 years ago

The answer to an addition problem is called a

Mathematics

2 answers:

Maslowich3 years ago

3 0

The answer to an addition problem is called a sum

Hope this helps!

Dennis_Churaev [7]3 years ago

3 0

Answer:

sum

Step-by-step explanation:

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In two or three sentences explain how you would solve for the real solutions of the following equation: please help asap!!

g100num [7]

the real solutions for the equation $x^{3}=-64$ are -

$x=4,2+-2\sqrt{3i}$

Step-by-step explanation:

$x^{3}$ = $- 64$

$x^{3} +64$ = 0

We can write 64 as $4^{3}$

$x^{3}$ + $4^{3}$ = 0

using the identity ( $x^{3}+y^{3} = (x+y)(x^{2} -xy+y^{2} )$ )

we get,

= $(x+4) (x^{2} -x*4+4^{2} )$

= $(x+4)(x^{2} -4x+16)$ ....................(1)

solving the quadratic equation ,

$x^{2} -4x+16$ =0

solutions of this quadratic equation can be obtained by

$x=-b +- \sqrt{b^{2}-4ac } /2a$

let use y for factors

$x=-(-4x)+-\sqrt{(-4x^{2} )-4*x^{2} *16} / 2*x^{2}$

$x=4x+-\sqrt{16x^{2} -64x^{2} } /2x^{2}$

$x=4+-\sqrt{16-64}/2$

$x=4+-4\sqrt{3i} /2$

<u /> $x=2+-2\sqrt{3i}$ ..................(2)

from the equation 1 we have,

$x-4=0$

which gives solution $x=4$

and from equation 2 we got $x=2+-2\sqrt{3i}$

so the real solutions for the equation $x^{3}=-64$ are -

$x=4,2+-2\sqrt{3i}$

3 0

2 years ago

PLEASE HELP!!!!!!!!!!!!1

skelet666 [1.2K]

A. Let x = cheese and
y = chocolate
2x + y = 25
x + y = 20

B. Subtract the second equation from the first.
2x + y = 25
-(x + y = 20)
-—————
x = 5

Plug 5 back in to the second equation and solve for y.
x + y = 20
5 + y = 20
Subtract 5 from both sides.
y = 15
5 cheese and 15 chocolate

Used elimination method because coefficients on the y values were both 1 so it was easy to subtract the equations and eliminate the y variable.

5 0

3 years ago

Billy’s Boat Rentals charges a yearly fee of $105 and $9.50 each time a member wants to rent a boat. Lisa’s Boat Rentals does no

AlekseyPX

Answer:

20 times

Step-by-step explanation:

The equation is...

9.5x+105=14.75x

105=5.25x

20=x

5 0