Tyler concludes that 5x² will always have a larger output for the same value of x.
<u>Look at the graph below and the table given</u>
Take a random value: x = 0
Here, 1 > 0, making 2^x > 5x²
Hence, 2^x is greater than 5x² at this point. making Tyler's point not applicable.
Disagree with Tyler's point.
Answer:
w=67°
Step-by-step explanation:
First you solve for x. 12x+11=x. X would be 1. Then you plug in 1 for x and add the w to the equation. 12(1)+11+w=90. after solving that you will get the total for w. which should be 67°
Answer:
okey???????????????????????? p and q is true got it.
Step-by-step explanation:
Close. You correctly set up the integrals. When integrating e²ˣ:
∫ e²ˣ dx
½ ∫ 2 e²ˣ dx
½ e²ˣ + C
So the coefficient should be ½, not 2.
[eˣ − ½ e²ˣ]₋₁⁰ + [½ e²ˣ − eˣ]₀¹
[(e⁰ − ½ e⁰) − (e⁻¹ − ½ e⁻²)] + [(½ e² − e) − (½ e⁰ − e⁰)]
1 − ½ − e⁻¹ + ½ e⁻² + ½ e² − e − ½ + 1
-e⁻¹ + ½ e⁻² + ½ e² − e + 1