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3 years ago

Simplify xz^3 x 4x^4z^4

Mathematics

1 answer:

beks73 [17]3 years ago

6 0

Answer:

4x^5z^7

Step-by-step explanation:

Multiply the whole numbers and add the exponents with the same base.

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8+5n is greater then or equal to 4 (n+8)

Semmy [17]

It would depend on what n was equal to

8 0

3 years ago

Read 2 more answers

7x/3 < 2 Pls answer asp please and thank you

d1i1m1o1n [39]

Answer:

x<6/7

Step-by-step explanation:

7x/3 < 2

7x < 6

x < 6/7

<h3>If you need to ask any questions, please let me know.</h3>

8 0

3 years ago

A tank contains 3,000 L of brine with 15 kg of dissolved salt. Pure water enters the tank at a rate of 30 L/min. The solution is

Shalnov [3]

Answer:

Step-by-step explanation:

Volume of tank is 3000L.

Mass of salt is 15kg

Input rate of water is 30L/min

dV/dt=30L/min

Let y(t) be the amount of salt at any time

Then,

dy/dt = input rate - output rate.

The input rate is zero since only water is added and not salt solution

Now, output rate.

Concentrate on of the salt in the tank at any time (t) is given as

Since it holds initially holds 3000L of brine then the mass to volume rate is y(t)/3000

dy/dt= dV/dt × dM/dV

dy/dt=30×y/3000

dy/dt=y/100

Applying variable separation to solve the ODE

1/y dy=0.01dt

Integrate both side

∫ 1/y dy = ∫ 0.01dt

In(y)= 0.01t + A, .A is constant

Take exponential of both side

y=exp(0.01t+A)

y=exp(0.01t)exp(A)

exp(A) is another constant let say C

y(t)=Cexp(0.01t)

The initial condition given

At t=0 y=15kg

15=Cexp(0)

Therefore, C=15

Then, the solution becomes

y(t) = 15exp(0.01t)

At any time that is the mass.

5 0

3 years ago

Sarah is a computer engineer and manager and works for a software company. She receives a

daser333 [38]

Answer:

a) Number of projects in the first year = 90

b) Earnings in the twelfth year = $116500

Total money earned in 12 years = $969000

Step-by-step explanation:

Given that:

Number of projects done in fourth year = 129

Number of projects done in tenth year = 207

There is a fixed increase every year.

a) To find:

Number of projects done in the first year.

This problem is nothing but a case of arithmetic progression.

Let the first term i.e. number of projects done in first year = $a$

Given that:

$a_4=129\\a_{10}=207$

Formula for $n^{th}$ term of an Arithmetic Progression is given as:

$a_n=a+(n-1)d$

Where $d$ will represent the number of projects increased every year.

and $n$ is the year number.

$a_4=129=a+(4-1)d \\\Rightarrow 129=a+3d .....(1)\\a_{10}=207=a+(10-1)d \\\Rightarrow 207=a+9d .....(2)$

Subtracting (2) from (1):

$78 = 6d\\\Rightarrow d =13$

By equation (1):

$129 =a+3\times 13\\\Rightarrow a =129-39\\\Rightarrow a =90$

Number of projects in the first year = 90

b)

Number of projects in the twelfth year =

$a_{12} = a+11d\\\Rightarrow a_{12} = 90+11\times 13 =233$

Each project pays $500

Earnings in the twelfth year = 233 $\times$ 500 = $116500

Sum of an AP is given as:

$S_n=\dfrac{n}{2}(2a+(n-1)d)\\\Rightarrow S_{12}=\dfrac{12}{2}(2\times 90+(12-1)\times 13)\\\Rightarrow S_{12}=6\times 323\\\Rightarrow S_{12}=1938$

It gives us the total number of projects done in 12 years = 1938

Total money earned in 12 years = 500 $\times$ 1938 = $969000

8 0

2 years ago

All I need is the answer. Will be marked brainiest if it is absolutely correct.

NARA [144]

What’s the question?

6 0

3 years ago