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3 years ago

How do you simplify numeric and algebraic expressions that contain exponents?

Mathematics

2 answers:

schepotkina [342]3 years ago

8 0

Simply simplify multiplication expressions with a positive exponent

maria [59]3 years ago

8 0

Answer:

just times it.

Step-by-step explanation:

hope it help

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How many miles is mars from earth

xxTIMURxx [149]

Answer:

33.9 million miles.

Step-by-step explanation:

4 0

3 years ago

Saul decides to use the IQR to measure the spread of the data. Saul calculates the IQR of the data set to be 27

german

Answer:

n = 78

Step-by-step explanation:

IQR is the interquartile range of the data set.

So Saul had already arranged the data in order, that is the first step when calculating the interquartile range.

25, 30, 50, 50, 50, 50, 56, n., 250

Next, is to find the median so we could easily separate the data into quartiles.

The median of the data 25, 30, 50, 50, 50, 50, 56, n., 250 = 50

Then, arrange the data into the first quartile and the third quartile

first quartile = 25, 30, 50, 50

third quartile = 50, 56, n., 250

first quartile median = 30 + 50 = 80/2 = 40

third quartile = 56 + n /2

The interquartile range formula is the first quartile subtracted from the third quartile:

IQR = Q3 – Q1.

Since 27 is the interquartile range, n=

27 = (56 + n /2) - 40

Use 2 to multiply through to eliminate the denominator

54 = (56 + n) - 80

n = 78

5 0

3 years ago

Please help me ASAP

stellarik [79]

Answer:

<em>The Graph is shown below</em>

Step-by-step explanation:

<u>The Graph of a Function</u>

Given the function:

$\displaystyle y=g(x)=-\frac{3}{2}(x-2)^2$

It's required to plot the graph of g(x). Let's give x some values:

x={-2,0,2,4,6}

And calculate the values of y:

$\displaystyle y=g(-2)=-\frac{3}{2}(-2-2)^2=-\frac{3}{2}(-4)^2==-\frac{3}{2}*16=-24$

Point (-2,-24)

$\displaystyle y=g(0)=-\frac{3}{2}(0-2)^2=-\frac{3}{2}(-2)^2=-\frac{3}{2}*4=-6$

Point (0,-6)

$\displaystyle y=g(2)=-\frac{3}{2}(2-2)^2=-\frac{3}{2}(0)^2=0$

Point (2,0)

$\displaystyle y=g(4)=-\frac{3}{2}(4-2)^2=-\frac{3}{2}(2)^2=-\frac{3}{2}*4=-6$

Point (4,-6)

$\displaystyle y=g(6)=-\frac{3}{2}(6-2)^2=-\frac{3}{2}(4)^2=-\frac{3}{2}*16=-24$

Point (6,-24)

The graph is shown in the image below

8 0

2 years ago

In a drawer, there are 3 white socks, 4 black socks, and 2 brown socks. You pick out a sock, replace or, turn pick put a 2nd soc

slega [8]

I believe the answer is B) 6/81

6 0

2 years ago

Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

Lina20 [59]

Answer:

The differential equation for the amount of salt A(t) in the tank at a time t > 0 is $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

Step-by-step explanation:

We are given that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

The concentration of the solution entering is 4 lb/gal.

Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;

$\frac{dA}{dt}= \text{R}_i_n - \text{R}_o_u_t$

where, $\text{R}_i_n$ = concentration of salt in the inflow $\times$ input rate of brine solution

and $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

So, $\text{R}_i_n$ = 4 lb/gal $\times$ 3 gal/min = 12 lb/gal

Now, the rate of accumulation = Rate of input of solution - Rate of output of solution

= 3 gal/min - 2 gal/min

= 1 gal/min.

It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.

So, $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

= $\frac{A(t)}{500+t} \text{ lb/gal } \times 2 \text{ gal/min}$ = $\frac{2A(t)}{500+t} \text{ lb/min }$

Now, the differential equation for the amount of salt A(t) in the tank at a time t > 0 is given by;

= $\frac{dA}{dt}=12\text{ lb/min } - \frac{2A(t)}{500+t} \text{ lb/min }$

or $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

4 0

3 years ago