Question

Medal for HELP :

Answer 1

The second question is it as two solutions.

Answer 2

Hmm you think so

Answer 3

yes, because it shows y=x-2 and -x+y=-5 and since there are two, there are two solutions.

Answer 4

it could be one though

Answer 5

since there are two equal signs it represents that there are two..that's what i recently learned.

Answer 6

alright y can you help me with the first one

Answer 7

I think it would be the first one

Answer 8

I dont think hes right Michele

Answer 9

but he might be with the first one

Answer 10

we can write this:
\[a = c \times \cos B\]

Answer 11

so we have:
\[\Large c = \frac{a}{{\cos B}} = \frac{a}{{\sqrt {1 - {{\left( {\sin B} \right)}^2}} }} = ...?\]

Answer 12

substituting our data, we get:
\[\Large c = \frac{a}{{\cos B}} = \frac{a}{{\sqrt {1 - {{\left( {\sin B} \right)}^2}} }} = \frac{{20}}{{\frac{{\sqrt 3 }}{2}}} = \frac{{40}}{{\sqrt 3 }}\]

Answer 13

now we have to rationalize. Please multiply both numerator and denominator by sqrt(3), what do you get?

Answer 14

I got 360

Answer 15

which cant be right

Answer 16

second question:
If I substitute the first equation into the second one, I get:
-x+x-2=-5
please simplify that expression

Answer 17

you get no solution

Answer 18

For the first one though I got 40

Answer 19

first question:
\[c = \frac{a}{{\cos B}} = \frac{a}{{\sqrt {1 - {{\left( {\sin B} \right)}^2}} }} = \frac{{20}}{{\frac{{\sqrt 3 }}{2}}} = \frac{{40}}{{\sqrt 3 }} = \frac{{40\sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }} = ...?\]

Answer 20

\[\large c = \frac{a}{{\cos B}} = \frac{a}{{\sqrt {1 - {{\left( {\sin B} \right)}^2}} }} = \frac{{20}}{{\frac{{\sqrt 3 }}{2}}} = \frac{{40}}{{\sqrt 3 }} = \frac{{40\sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }} = ...?\]

Answer 21

I get because i combined them

Answer 22

that's right!

Answer 23

sweet

Answer 24

hint:
second question, after a simplification we have:
-2=-5

Answer 25

well then two solutions then -2=-5

Answer 26

no, it is not an equation, since the statement
-2=-5 is false, so we have no solutions

Answer 27

oh I thought you meant i was wrong... sorry my bad

Answer 28

Thank you for explaining those to me!

Answer 29

Thank you!