How do you graph y=2(1/3)^-2

PLEASE HELP QUICKLY IF YOU CAN!

Anastasy [175]

Answer:

6 - $(\frac{2}{3} )^{2} =$ $\frac{50}{9}$

3² - $(\frac{2}{3} )=$ $8\frac{1}{3}$

7 - $(\frac{1}{2} )^{3}$ = $\frac{55}{8}$

Step-by-step explanation:

1. 6 - $(\frac{2}{3} )^{2}$

First, apply the exponent to 2/3. Because it's to the second power, multiply 2/3 by itself.

$(\frac{2}{3} )^{2}=\frac{4}{9}$

6 - $\frac{4}{9}$

Now, subtract.

6 - $\frac{4}{9}$ = $\frac{50}{9}$

------------------------------------------

2. 3² - $\frac{2}{3}$

First, apply the exponent.

3² = 9

9 - $\frac{2}{3}$

Now, subtract.

9 - $\frac{2}{3}=\frac{25}{3}$

$\frac{25}{3} =8\frac{1}{3}$

------------------------------------------

3. 7 - $(\frac{1}{2} )^{3}$

First, apply the exponent.

$(\frac{1}{2} )^{3}$ = $\frac{1}{8}$

7 - $\frac{1}{8}$

Now, subtract.

7 - $\frac{1}{8}=\frac{55}{8}$

hope this helps!

4 0

2 years ago

An amount of R351 is made up with 114 coins (R2 coins and R5 coins only). How many R2 coins are there? How many R5 coin are ther

lisabon 2012 [21]

Answer:

41

Step-by-step explanation:

For this, you have to make two equations;

One that calculates the amount of coins

and one that calculates the amount of R

I'm going to set

R2 = x

R5 = y

To make this a bit easier to look at

So to do the number of coins, you want to do this:

x + y = 114

This is saying that the total amount of R2 coins plus the total amount of R5 coins will add up to 114

Then, you want to calculate the worth of them.

2(x) + 5(y) = 351

This is saying that for ever R2 coin there is, it will increase the total by 2, and every R5, it will increase the total by 5. When you add these together, it will come out to 351.

Then you have to solve the system of equations:

x + y = 114

2x + 5y = 351

-2 * (x + y = 114)

2x + 5y = 351

-2x + -2y = -228

2x + 5y = 351

3y = 123

y = 41

Since y is the amount of R5 coins, we know we have 41 R5 coins.

So the answer is

41

4 0

3 years ago

Today is the 65th day of school. If we are out of 180 days, how do I write this as a

jeka57 [31]

13/36 would be the answer I believe

8 0

3 years ago

How many outcome sequences are possible when a die is rolled 4 times where we say for instance that the outcome is 3 4 3 1 if th

STatiana [176]

Answer:

the possible outcome sequences when a die is rolled 4 times is 1296

Step-by-step explanation:

Given the data in the question;

a die is rolled 4 times

and outcomes are { 3, 4, 3, 1 }

we know that; possible number of outcomes on a die is n = 6{ 1,2,3,4,5,6 }

Now when we roll a die lets say, r times

then the total number of possible outcomes will be;

N = $n^r$

given that; r = 4

Hence if we roll a die 4 times;

Total number of possible outcome N = 6⁴

N = 1296

Therefore, the possible outcome sequences when a die is rolled 4 times is 1296

3 0

3 years ago

You fit a trendline of y = 2x 1 to your data. what would the residual be for a data point measured as y = 3.5 for an x value of

nata0808 [166]

For a data point measured as y = 3.5 for an x value of x = 1.2, the residual would be 0.1

For given question,

we have been given an equation 2x + 1 that represents an equation for a trendline to the data.

General formula with residual is y = 2x + 1 + r, where r is a residual.

We need to find the residual for a data point measured as y = 3.5 for an x value of x = 1.2

We substitute given values of x and y in the residual equation

y = 2x + 1 + r

For y = 3.5 and x = 1.2,

⇒ 3.5 = 2(1.2) + 1 + r

⇒ 3.5 = 2.4 + 1 + r

⇒ r = 3.5 - 3.4

⇒ r = 0.1

Therefore, for a data point measured as y = 3.5 for an x value of x = 1.2, the residual would be 0.1

Learn more about the residual here:

brainly.com/question/10228420

#SPJ4

7 0

2 years ago