Part A.

Which equation is equivalent to 16 Superscript 2 p Baseline = 32 Superscript p 3?.

Mrac [35]

The equation which is equivalent to 16 Superscript 2 p Baseline equal to 32 Superscript p 3 is,

$2^{8p}=2^{5p+15}$

<h3>What is equivalent equation?</h3>

Equivalent equation are the expression whose result is equal to the original expression, but the way of representation is different.

Given information-

The given equation in the problem is,

$16^{2p}=32^{p+3}$

Write both the equation in the form of same base number as,

$(2^4)^{2p}=(2^5)^{p+3}$

The power of the power of a number can be written as product of both the numbers. Thus,

$(2)^{4\times2p}=(2)^{5\times(p+3)}\\2^{8P}=2^{5P+15}$

This is the required equation.

Now if the base is the same at both side of the expression, then the powers can be compared. Thus,

$8p=5p+15$

Solve it further to find the value of p as,

$8p-5p=15\\3p=15\\p=5$

Thus the equation which is equivalent to 16 Superscript 2 p Baseline equal to 32 Superscript p 3 is,

$2^{8p}=2^{5p+15}$

Learn more about the equivalent expression here;

brainly.com/question/2972832

7 0

2 years ago

How do I work this out

Iteru [2.4K]

If the track is 10km long, and X travels at 40 km/h, it will take 1 hour to do 4 laps. this means that car X will do one lap in 15 minutes. Y travels at 30 km/h, which means it will do 3 laps in 1 hour, meaning that car X will do 1 lap in 20 minutes.

X- 15, 30, 45, 60
Y-20, 40, 60

Therefore, after an hour, they will pass again.

5 0

3 years ago

-8x - 8y = 0 and -5x - 4y = 5. Solve the system of equations

Rashid [163]

Answer:

x=-5, y=5 or (-5,5)

Step-by-step explanation:

We can do elimination with this

this means we should first multiply -5x-4y=5 by 2

this would make it -10x-8y=10

we can now subtract this from the original equation

-8x-8y=0 - (-10x-8y=10)

the ys would cancel out

so this would equal

2x=-10

x=-5

now we can plug this back into the equation to find y

-8(-5)-8y=0

40-8y=0

-8y=-40

y=5

4 0

3 years ago

Read 2 more answers

Integration of dx/3sin^2 -4

Scilla [17]

Answer:

= 1/2 tan^(-1)(2 cot(x)) + constant

Step-by-step explanation:

Take the integral:

integral1/(3 sin^2(x) - 4) dx

Multiply numerator and denominator of 1/(3 sin^2(x) - 4) by -sec^2(x):

= integral-(sec^2(x))/(4 sec^2(x) - 3 tan^2(x)) dx

Prepare to substitute u = tan(x). Rewrite -(sec^2(x))/(4 sec^2(x) - 3 tan^2(x)) using sec^2(x) = tan^2(x) + 1:

= integral-(sec^2(x))/(tan^2(x) + 4) dx

For the integrand -(sec^2(x))/(tan^2(x) + 4), substitute u = tan(x) and du = sec^2(x) dx:

= integral-1/(u^2 + 4) du

Factor out constants:

= - integral1/(u^2 + 4) du

Factor 4 from the denominator:

= - integral1/(4 (u^2/4 + 1)) du

Factor out constants:

= -1/4 integral1/(u^2/4 + 1) du

For the integrand 1/(u^2/4 + 1), substitute s = u/2 and ds = 1/2 du:

= -1/2 integral1/(s^2 + 1) ds

The integral of 1/(s^2 + 1) is tan^(-1)(s):

= -1/2 tan^(-1)(s) + constant

Substitute back for s = u/2:

= -1/2 tan^(-1)(u/2) + constant

Substitute back for u = tan(x):

= -1/2 tan^(-1)(tan(x)/2) + constant

Which is equivalent for restricted x values to:

Answer: = 1/2 tan^(-1)(2 cot(x)) + constant

6 0

3 years ago

Solve the quadratic equation numerically (using tables of x- and y- values).

garri49 [273]

benceA yada D bilemedim

6 0

3 years ago