All categories

3 years ago

I need help Simplify. (x³)⁵

Mathematics

1 answer:

pshichka [43]3 years ago

7 0

The answer would be X^15

You might be interested in

Write the integer from least to greatest. 4, 12, -12, -11

maxonik [38]

Answer:

-12, -11, 4, 12

Step-by-step explanation:

A negative number is less the higher it is, and a positive number is more the higher it is

3 0

2 years ago

1500/semiannual period for 6 years at 2.5%/year compounded semiannually

kari74 [83]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$1500\\ r=rate\to 2.5\%\to \frac{2.5}{100}\dotfill &0.025\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{semiannual, thus twice} \end{array}\dotfill &2\\ t=years\dotfill &6 \end{cases} \\\\\\ A=1500\left(1+\frac{0.025}{2}\right)^{2\cdot 6}\implies A=1500(1.0125)^{12}\implies A\approx 1741.13$

6 0

2 years ago

A boat makes a 120-mile trip downstream in 3 hours but makes the return trip in 4 hours.

Free_Kalibri [48]

Answer: C 5mph

Hope this helps!!

6 0

3 years ago

Suppose that F(x)= x^2 and G(x) 4/5 x^2. Which statements best compares the graph of G(x) with the graph of F(x)?

34kurt

Answer:

Step-by-step explanation:

C. The graph of G(x) is the graph of F(x) compressed vertically.

If, for example, x = 1, then F(1) = 1^2 = 1, while G(1) = (4/5)(1^2) = 4/5.

Note that G(1) < F(1); we say G(x) is a vertical compression of F(x).

8 0

2 years ago

Suppose that a large mixing tank initially holds 900 gallons of water in which 50 pounds of salt have been dissolved. Another br

borishaifa [10]

Answer:

dA/dt= 12-2A(t)/900+t

Step by step Explanation:

Given:

rate of 3 gal/min

concentration of the solution entering is 4 lb/gal,

The equation of the solution can be expressed as

dA/dt = R(1)- R(2)

Where R(1)=( concentration of the salt at inflow* rate of input of given brine)

R(2)= ( concentration of the salt at inflow* rate of output of given brine)

R(1)= 3 gal/min× 4 lb/gal=12Ib/min

But the solution is been pumped slowly, the rate of the accumulation can calculated as;

(3 gal/min- 2 gal/min)=1gal/min

But after t min we will be having 900+t gallons present in the tank

CHECK THE ATTACHMENT FOR COMPLETE EXPLANATION

7 0

2 years ago