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3 years ago

MORE FREEEEEE POINTSSSSSSSSS!!!

Mathematics

2 answers:

vova2212 [387]3 years ago

6 0

Ily ❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️ have a good day

Romashka-Z-Leto [24]3 years ago

3 0

Answer:

yoyooyyoyoyoyoyyoyoyoyoyoyoyo

Step-by-step explanation:

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Use abc to find the value of cos B

krok68 [10]

Cos of b equals c over a

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3 years ago

PLEASE HELP!!<br><br> Question in picture below

Oksanka [162]

Step-by-step explanation:

Absolute error = 300-286

=14 pounds

Percentage error = 14/286 x 100%

=4.8951

=4.9%

4 0

3 years ago

Read 2 more answers

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 -

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

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2 years ago

*EXTRA PTS* answer #2 - #5

TiliK225 [7]

Answer:

2. 2N

3. -4N

4. 4N

5. 7N

Step-by-step explanation:

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3 years ago

Which of these equations represent functions? Check all that apply.

netineya [11]

The answer to the question is D

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3 years ago