Yeah it would be A. because x approaching infinity is to the right of the graph, and since there is a horizontal asymptote at -3 then its correct
1.) add 9 and 5 which is 14-6x
Answer:
(D) t^4
Step-by-step explanation:
You have defined ...
a3 = a2·a1
a4 = a3·(a2·a1) = a3²
a5 = a4·(a3·a2·a1) = a4² = (a3²)² = a3⁴
Then if a3 = t, a5 = t⁴
Answer/Step-by-step explanation:
-33 is not a whole number. A whole number is a number(positive) that contains no decimals nor fractions.
Answer:
Step-by-step explanation:
|4x-3|=5√(x+4) ⇔ |4x-3|²=5²(√(x+4))² and x+4 ≥ 0
⇔ (4x-3)² = 25(x+4) and x+4 ≥ 0 ( because : /a/² = a²)
⇔16x²-24x+9 = 25x +100 and x+4 ≥ 0
⇔ 16x² -49x - 91 =0 and x+4 ≥ 0 quadratic equation
Δ = (-49)²-4(16)(-91) = 8225
two solution : X1 = (49-√8225)/32 ≅ - 1.3 accept (-1.3+4 ≥ 0)
X2 = (49+√8225)/32 ≅4.37 accept (4.37+4 ≥ 0)
