All categories

4 years ago

Can someone please help!! (Simplifying exponential equations)

Mathematics

1 answer:

Snezhnost [94]4 years ago

6 0

Seems we undistributed the $( \frac{5}{2} )^x$

remember
$x^{a+b}=(x^a)(x^b)$
factor out $( \frac{5}{2} )^x$ from each term on the left side
$1( \frac{5}{2} )^x+( \frac{5}{2} )^3( \frac{5}{2} )^x=A( \frac{5}{2} )^x$
undistribute $( \frac{5}{2} )^x$
$( \frac{5}{2} )^x(1+( \frac{5}{2} )^3)=A( \frac{5}{2} )^x$
divide both sides by $( \frac{5}{2} )^x$
$1+( \frac{5}{2} )^3=A$

You might be interested in

A population has a mean of 400 and a standard deviation of 90. suppose a sample of size 100 is selected and x is used to estimat

Over [174]

What were your given answer choices

5 0

4 years ago

Explain how the Quotient of Powers was used to simplify this expression. (NEED HELP ASAP, 20 POINTS!)

taurus [48]

The answer to this question will be C

7 0

4 years ago

A child wanders slowly down a circular staircase from the top of a tower. With x,y,zx,y,z in feet and the origin at the base of

babymother [125]

Answer:

a) The tower is 90 feet tall

b) She reaches the bottom at t = 18 minutes.

c) Her speed at time t is 5 \sqrt[]{5} ft/minute

d) Her acceleration at time t is 10 ft/minute^2

Step-by-step explanation:

Consider the path described by the child as going down the tower to have the following parametrization $\gamma(t) = (10\cos t, 10 \sin t, 90-5t)$

a) Assuming that the child is at the top of the tower when she starts going down, we have that at the initial time (t=0) we will have the value of the height of the tower. That is z = 90-5*0 = 90 ft.

b) The child reaches the bottom as soon as z =0. We want to find the value of t that does that. Then we have 0 = 90-5t, which gives us t = 18 minutes.

c) Given the parametrization we are given, the velocity of the child at time t is given by $\frac{d\gamma}{dt}= (\frac{d}{dt}(10\cos t), \frac{d}{dt} (10 \sin t ), \frac{d}{dt}(90-5t)) = (-10 \sin t, 10 \cos t, -5)$ . The speed is defined as the norm of the velocity vector,

so, the speed at time t is given by $v = \sqrt[]{(-10 \sin t)^2+(10 \cos t)^2+(-5)^2} = \sqrt[]{100(\sin^2 t + \cos^2 t)+25} = \sqrt[]{125}= 5 \sqrt[]{5}$

d) ON the same fashion we want to know the norm of the second derivative of $\gamma$ .

We have that $\gamma ^{''}(t) =(-10\cost t, -10 \sin t , 0)$ so the acceleration is given by $\sqrt[]{100(\cos^2 t+ \sin^2 t )} = 10$

6 0

3 years ago

A can of beans has surface area 320cm squared . Its height is 14 cm. What is the radius of the circular top?

Ratling [72]

Steps:

All cans take on the shape of a cylinder, unless you have seen interesting shape of cans like a starfish.

The formula for surface area of a cylinder is

SA = 2πr2 + 2πrh

where:

r = radius

h = height

Since we know the surface area and height, we can plug them in. Note that we can factor out the 2π. You will see why we factor out 2π rather than 2πr.

2π(r2 + (20)r) = 396

2π(r2 + 20r) = 396

Divide both sides of the equation by 2π to isolate the r terms.

r2 + 20r = 63.025

Subtract 63.025 on both sides of the equation.

r2 + 20r - 63.025 = 0

Use the quadratic formula to solve for r:

r = (-b ± √(b2 - 4ac)) / 2a

where:

a = 1

b = 20

c = -63.025

Plug in these values into the formula. You should get two solutions because of the plus/minus sign. Accept the positive value of r.

Please mark brainliest

<em><u>Hope this helps.</u></em>

4 0

3 years ago

On one of its routes across Asia, Alpha Airlines flies an aircraft with checked-in luggage capacity of 8500 lbs. There are 121 s

Doss [256]

Answer:

the probability is P=0.012 (1.2%)

Step-by-step explanation:

for the random variable X= weight of checked-in luggage, then if X is approximately normal . then the random variable X₂ = weight of N checked-in luggage = ∑ Xi , distributes normally according to the central limit theorem.

Its expected value will be:

μ₂ = ∑ E(Xi) = N*E(Xi) = 121 seats * 68 lbs/seat = 8228 lbs

for N= 121 seats and E(Xi) = 68 lbs/person* 1 person/seat = 68 lbs/seat

the variance will be

σ₂² = ∑ σ² (Xi)= N*σ²(Xi) → σ₂ = σ *√N = 11 lbs/seat *√121 seats = 121 Lbs

then the standard random variable Z

Z= (X₂- μ₂)/σ₂ =

Zlimit= (8500 Lbs - 8228 lbs)/121 Lbs = 2.248

P(Z > 2.248) = 1- P(Z ≤ 2.248) = 1 - 0.988 = 0.012

P(Z > 2.248)= 0.012

then the probability that on a randomly selected full flight, the checked-in luggage capacity will be exceeded is P(Z > 2.248)= 0.012 (1.2%)

8 0

3 years ago