Please help ...................!

wariber [46]

True, Fe (iron) has 2 on the left side and on the other side as well. C (carbon) has 3 on the left and 3 on the other. O (oxygen) has 6, 3 for the first part and another 3 because of the reciprocal in the beginning of the equation. On the other side O has 2 and it's multiplied by the reciprocal 3 which is 6

3 0

3 years ago

How many different 3-digit PIN codes are there that only include the digits 7, 4, 3 and 2?

KengaRu [80]

Answer:

64

Step-by-step explanation:

We can use the product rule of counting to answer this:

There are 4 possibilities for the 1st digit
There are 4 possibilities for the 2nd digit
There are 4 possibilities for the 3rd digit
So we use the product rule of counting to give us:

4×4×4=64

6 0

3 years ago

Help with thissssssss

kompoz [17]

Answer:

27

Step-by-step explanation:

1^3 = 1

2^3 = 8

3^3 = 27

4^3 = 64

5^3 = 125

The missing number is 27.

3 0

3 years ago

Read 2 more answers

Tamira invests $5,000 in an account that pays 4% annual interest. How much will there be in the account after 3 years if the int

Talja [164]

Answer:

There will be $5624.32 in the account after 3 years if the interest is compounded annually.

There will be $5630.812 in the account after 3 years if the interest is compounded semi-annually.

There will be $5634.125 in the account after 3 years if the interest is compounded quarterly.

There will be $5636.359 in the account after 3 years if the interest is compounded monthly

Step-by-step explanation:

Tamira invests $5,000 in an account

Rate of interest = 4%

Time = 3 years

Case 1:

Principal = 5000

Rate of interest = 4%

Time = 3 years

No. of compounds per year = 1

Formula : $A=P(1+r)^t$

$A=5000(1+0.04)^3$

A=5624.32

There will be $5624.32 in the account after 3 years if the interest is compounded annually.

Case 2:

Principal = 5000

Rate of interest = 4%

Time = 3 years

No. of compounds per year = 2

Formula : $A=P(1+\frac{r}{n})^{nt}$

$A=5000(1+\frac{0.04}{2})^{2 \times 3}$

A=5630.812

There will be $5630.812 in the account after 3 years if the interest is compounded semi-annually.

Case 3:

Principal = 5000

Rate of interest = 4%

Time = 3 years

No. of compounds per year = 4

Formula : $A=P(1+\frac{r}{n})^{nt}$

$A=5000(1+\frac{0.04}{4})^{4 \times 3}$

A=5634.125

There will be $5634.125 in the account after 3 years if the interest is compounded quarterly.

Case 4:

Principal = 5000

Rate of interest = 4%

Time = 3 years

No. of compounds per year = 4

Formula : $A=P(1+\frac{r}{n})^{nt}$

$A=5000(1+\frac{0.04}{12})^{12 \times 3}$

A=5636.359

There will be $5636.359 in the account after 3 years if the interest is compounded monthly

8 0

4 years ago

S see Use the graph y=e^x to evaluate these expression e^0

nika2105 [10]

Answer:

$y= e^x$

And we want to evaluate e^0 using the graph

And as we can see in the plot the y intercept is the blue point with y=0 and that correspond with:

$y(0)= e^0 =1$

Step-by-step explanation:

For this problem we know the following function:

$y= e^x$

And we want to evaluate e^0 using the graph

And as we can see in the plot the y intercept is the blue point with y=0 and that correspond with:

$y(0)= e^0 =1$

5 0

3 years ago