Answer:
<em>t</em><em>h</em><em>e</em><em> </em><em>c</em><em>o</em><em>r</em><em>r</em><em>e</em><em>c</em><em>t</em><em> </em><em>a</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em> </em><em>i</em><em>s</em><em> </em><em>C</em><em> </em><em>1</em><em>/</em><em>3</em><em>2</em><em> </em><em>I</em><em>b</em>
Step-by-step explanation:
hope it helps
Answer:
you domb if you dont know the answer B
Step-by-step explanation:
Subtract 9x from 23x yields 14x. This was done by subtracting algebraically the variables.
Answer:
Step-by-step explanation:
Given the expression cosec (x) = 4 and tan(x)< 0
since cosec x = 1/sinx
1/sinx = 4
sinx = 1/4
From SOH, CAH TOA
sinθ = opposite/hypotenuse
from sinx = 1/4
opposite = 1 and hypotenuse = 4
to get the adjacent, we will use the Pythagoras theorem
adj² = 4²-1²
adj² = 16-1
adj ²= 15
adj = √15
cosx = adj/hyp = √15/4
tanx = opposite/adjacent = 1/√15
since tan < 0, then tanx = -1/√15
From double angle formula;
sin2x = 2sinxcosx
sin2x = 2(1/4)(√15/4)
sin2x = 2√15/16
sin2x = √15/8
for cos2x;
cos2x = 1-2sin²x
cos2x = 1-2(1/4)²
cos2x = 1-2(1/16)
cos2x= 1-1/8
cos2x = 7/8
for tan2x;
tan2x = tanx + tanx/1-tan²x
tan2x = 2tanx/1-tan²x
tan2x = 2(-1/√15)/1-(-1/√15)²
tan2x = (-2/√15)/(1-1/15)
tan2x = (-2/√15)/(14/15)
tan2x = -2/√15 * 15/14
tan2x = -30/14√15
tan2x = -30/7√15
rationalize
tan2x = -30/7√15 * √15/√15
tan2x = -30√15/7*15
tan2x = -2√15/7
54 divided by 6 would be 9 meaning he takes 9 quizzes a week. If directly proportional by the end of 7 weeks he would have taken 63 quizzes