In triangle $DEF,$ $\angle D = 30^\circ,$ $\angle F = 60^\circ,$ and $EF = 6.$ Find $DE + DF.$ [asy]
unitsize(2 cm);

pair D, E, F;

D = (0,sqrt(3));
E = (0,0);
F = (1,0);

draw(D--E--F--cycle);

label("$D$", D, N);
label("$E$", E, SW);
label("$F$", F, SE);
label("$6$", (E + F)/2, S);
[/asy]

Please no links!

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

well there are two answers for that.

1: you need 120 ones then that will equal up to 120.

2: you could also do one 100 and 2 tens.

those are the answers that i have thought of.

the inequality is:

250c+180g≥950

step-by-step explanation:

let c denotes the number of days in a week   miyoko worked as a cryptographer.

and g denotes the number of days in a week she works as a geologist.

now we are given that she earns $250 for every day she works as a cryptographer,and earns $180 for every day she works as a geologist.

now miyoko's goal is to earn more than $950 this week.

hence, the inequality could be represented as:

250c+180g≥950