Easy
recall that
an=a1(r)^(n-1)
so
given 2nd and 5th term
we get
a2 and a5
so
a2=a1(r)^(2-1)=a1(r)^1=a1r
a5=a1(r)^(5-1)=a1(r)^4
also remember that
so
so r^3=-64
cube root
r=-4
so
a2=a1r=-8
a2=a1(-4)=-8
divide both sides by -4
a1=2
so
equation is
C isi the answer
Answer:
Step-by-step explanation:
∠ADC = 90°
∠ADB + ∠BDC = 90
20 + ∠BDC = 90
∠BDC = 90 - 20 = 70
∠BDC = 70°
16) ∠PSR = ∠PSQ +∠QSR
= 60 + 10
= 70
18) ∠PSR = 130
∠PSQ + ∠QSR = 130
90+ ∠QSR = 130
∠QSR = 130 - 90
∠QSR = 40
19)∠ADC = ∠ADB + ∠BDC
= 120 +20
= 140
20) ∠PSR = 125
Answer:
18
Step-by-step explanation:
f(x) is the ultimate base of all function notation forms
f(h) is the addition of 4(Vertical Shift of 4)
f(h)=2x+10
If f(4), then 2(4)+10
f(4)=18
Answer:(a)x^2+2y^2=2
(b)In the attached diagram
Step-by-step explanation:Step 1: Multiply both equations by t
xt=t(cost -sint)\\ty\sqrt{2} =t(cost +sint)
Step 1: Multiply both equations by t
xt=t(cost -sint)\\ty\sqrt{2} =t(cost +sint)
Step 2:We square both equations
(xt)^2=t^2(cost -sint)^2\\(ty)^2(\sqrt{2})^2 =t^2(cost +sint)^2
Step 3: Adding the two equations
(xt)^2+(ty)^2(\sqrt{2})^2=t^2(cost -sint)^2+t^2(cost +sint)^2\\t^2(x^2+2y^2)=t^2((cost -sint)^2+(cost +sint)^2)\\x^2+2y^2=(cost -sint)^2+(cost +sint)^2\\(cost -sint)^2+(cost +sint)^2=2\\x^2+2y^2=2 hopes this helps
