Yes I know it’s to much but pleaseee...it’s due at 11 :(

Which is a zero of the quadratic function f(x) = 16x?? + 32x - 9?

STALIN [3.7K]

Answer:

f(x) = 16x2 + 32x - 9

To find the "zeros", set f(x) = 0:

0 = 16x2 + 32x - 9

The equation is solvable by factoring or by using the quadratic formula. The factors (use the "ac" method) are:

0 = (4x-1)(4x+9)

x = 1/4 = 0.25 and x = -9/4 = -2.25

Step-by-step explanation:

6 0

3 years ago

What is the area of a triangle with vertices at (-3 3) (-3,2) and (1,2)?

d1i1m1o1n [39]

This is a right triangle, so first find the distance between the two legs...

Points where the numbers are the same indicate a point that is on the same axis as the other

(-3, 3) and (-3, 2) have a distance of 1

(-3,2) and (1,2) have a distance of 4

The area formula for a triangle is 1/2bh and in this case 1/2(1)(4) = 2

The area is 2

6 0

2 years ago

Which of these is a verb? Europe Pretty Cup Glanced

Serjik [45]

Hello there! A verb is a word that takes an action of doing something, so the answer would be Glanced.
Hope this helps you!

7 0

3 years ago

The school cafeteria is holding an election for students to vote on which items they would like to see on the lunch menu. The ch

d1i1m1o1n [39]

There are two choices to choose from:

a) Grilled chicken
b) Veggie Pizza

36% students voted for veggie pizza. This means the remaining votes will go to the Grilled chicken. Thus grilled chicken received 64% of the student votes.

Since more students voted for the Grilled chicken, it will be on the lunch menu.

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%20%5C%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos2x%7D-%5Csqrt%5B3%5D%7Bcos3x%7D%20%7D%7

salantis [7]

Answer:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

We are given the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}$

When we directly plug in x = 0, we see that we would have an indeterminate form:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}$

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}$

Plugging in x = 0 again, we would get:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}$

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}$

Substitute in x = 0 once more:

$\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}$

And we have our final answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

6 0

3 years ago