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2 years ago

12.7, 12 3/5, 12 3/4 greatest to lest

Mathematics

1 answer:

horsena [70]2 years ago

6 0

Answer:

12 3/4, 12.7, 12 3/5

Step-by-step explanation:

12 3/5 = 12.60

12 3/4 = 12.75

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Use the distributive property to write an expression that is equivalent to each expression 1/5(20y - 4x - 13)

DiKsa [7]

The answer should be
4y-4/5x-13/5

8 0

2 years ago

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

How do I solve this question (19a)?

fredd [130]

Answer i think would be
l-4x+4l
---------
x-1

8 0

3 years ago

6x - 4 = 16

Juli2301 [7.4K]

Hello!

Answer:

x=10/3

Step-by-step explanation:

First you want start by adding 4 to both sides

6x-4+4=16+4

6x=20

Then divide 6 to both sides,

6x/6=20/6

x= 10/3

Hope this helps! :)

4 0

2 years ago

What’s the answer to this?

Elina [12.6K]

Step-by-step explanation:

Line is passing through the points:

$(-2,\:3)=(x_1,\:y_1) \:\&\: (2,\:5)=(x_2 ,\:y_2)$

Equation of line in two point form is given as:

$\frac{y -y_1 }{y_1 -y_2 } = \frac{x -x_1 }{x_1 -x_2 } \\ \\ \therefore \frac{y -3}{ 3 -5} = \frac{x -(-2) }{-2 -2} \\ \\ \therefore \frac{y -3}{ - 2} = \frac{x -(-2) }{-4} \\ \\ \therefore \frac{y - 3}{1} = \frac{x +2}{2} \\ \\ \therefore \: y-3= \frac{x}{2} + \frac{2}{2} \\ \\\therefore \: y= \frac{x}{2} + 1 +3\\ \\ \huge \red{ \boxed{\therefore \: y= \frac{1}{2} \:x+ 4}}\\ is \: the \: required \: equation \: of \: line.$

4 0

2 years ago