Simplify radical 125

Use a graphing calculator to solve the equation -3 cost= 1 in the interval from . Round to the nearest hundredth.

RSB [31]

In place of t, or theta, I'm going to utilize x instead. So the equation is -3*cos(x) = 1. Get everything to one side and we have -3*cos(x)-1 = 0

Let f(x) = -3*cos(x)-1. The goal is to find the root of f(x) in the interval [0, 2pi]

I'm using the program GeoGebra to get the task done of finding the roots. In this case, there are 2 roots and they are marked by the points A and B in the attachment shown

A = (1.91, 0)
B = (4.37, 0)

So the two solutions for theta are
theta = 1.91 radians
theta = 4.37 radians

8 0

3 years ago

Which strategies can be used to solve this problem?

vladimir2022 [97]

You want total number made in 4 m onths, so adding all numbers is correct, but it is per 1 month, so multiply by 2

A because you need brain for most problems
D works

pick A and D

8 0

3 years ago

First you to find the worksheet and download it<br> plase I need help

Goshia [24]

Answer:

a) The horizontal asymptote is y = 0

The y-intercept is (0, 9)

b) The horizontal asymptote is y = 0

The y-intercept is (0, 5)

c) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

d) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

e) The horizontal asymptote is y = -1

The y-intercept is (0, 7)

The x-intercept is (-3, 0)

f) The asymptote is y = 2

The y-intercept is (0, 6)

Step-by-step explanation:

a) f(x) = $3^{x + 2}$

The asymptote is given as x → -∞, f(x) = $3^{x + 2}$ → 0

∴ The horizontal asymptote is f(x) = y = 0

The y-intercept is given when x = 0, we get;

f(x) = $3^{0 + 2}$ = 9

The y-intercept is f(x) = (0, 9)

b) f(x) = $5^{1 - x}$

The asymptote is fx) = 0 as x → ∞

The asymptote is y = 0

Similar to question (1) above, the y-intercept is f(x) = $5^{1 - 0}$ = 5

The y-intercept is (0, 5)

c) f(x) = 3ˣ + 3

The asymptote is 3ˣ → 0 and f(x) → 3 as x → ∞

The asymptote is y = 3

The y-intercept is f(x) = 3⁰ + 3= 4

The y-intercept is (0, 4)

d) f(x) = 6⁻ˣ + 3

The asymptote is 6⁻ˣ → 0 and f(x) → 3 as x → ∞

The horizontal asymptote is y = 3

The y-intercept is f(x) = 6⁻⁰ + 3 = 4

The y-intercept is (0, 4)

e) f(x) = $2^{x + 3}$ - 1

The asymptote is $2^{x + 3}$ → 0 and f(x) → -1 as x → -∞

The horizontal asymptote is y = -1

The y-intercept is f(x) = $2^{0 + 3}$ - 1 = 7

The y-intercept is (0, 7)

When f(x) = 0, $2^{x + 3}$ - 1 = 0

$2^{x + 3}$ = 1

x + 3 = 0, x = -3

The x-intercept is (-3, 0)

f) $f(x) = \left (\dfrac{1}{2} \right)^{x - 2} + 2$

The asymptote is $\left (\dfrac{1}{2} \right)^{x - 2}$ → 0 and f(x) → 2 as x → ∞

The asymptote is y = 2

The y-intercept is f(x) = $f(0) = \left (\dfrac{1}{2} \right)^{0 - 2} + 2 = 6$

The y-intercept is (0, 6)

7 0

3 years ago

I need helps someone

umka21 [38]

The answer is 3/2,-6

5 0

3 years ago

Read 2 more answers

Given: mAngleEDF = 120°; mAngleADB = (3x)°; mAngleBDC = (2x)° Prove: x = 24 3 lines are shown. A line with points E, D, C inters

N76 [4]

Answer:

" Vertical angles are congruent " ⇒ 2nd answer

Step-by-step explanation:

* <em>Look to the attached figure </em>

- There are three lines intersected at point D

- We need to find the missing in step 3

∵ Line FA intersects line EC at point D

- The angles formed when two lines cross each other are called

vertical angles

- Vertical angles are congruent (vertical angles theorem)

∴ ∠ADC and ∠FDE are vertical angles

∵ Vertical angles are congruent

∴ ∠EDF ≅ ∠ADC

∴ m∠EDF ≅ m∠ADC

∵ m∠EDF = 120° ⇒ given

∵ m∠ADC = m∠ADB + m∠BDC

∴ m∠ADB + m∠BDC = 120°

∵ m∠ADB = (3x)° ⇒ given

∵ m∠BDC = (2x)° ⇒ given

∴ 3x + 2x = 120 ⇒ add like terms

∴ 5x = 120 ⇒ divide both sides by 5

∴ x = 24

Column (1) Column (2)

m∠EDF = 120° given

m∠ADB = 3 x given

m∠BDC = 2 x given

∠EDF and ∠ADC are vertical angles defin. of vert. ∠s

∠EDF is congruent to ∠ADC vertical angles are

congruent

m∠ADC = m∠ADB + m∠BDC angle add. post.

m∠EDF = m∠ADC defin. of cong.

m∠EDF = m∠ADB + m∠BDC substitution

120° = 3 x + 2 x substitution

120 = 5 x addition

x = 24 division

∴ The missing reason is " vertical angles are congruent "

- From the explanation above ∠ADC and ∠FDE are vertical

angles then they are congruent according to vertical angle

theorem

6 0

4 years ago

Read 2 more answers