Which of the following is the same as 8.4 × 10 to the 2nd power

How many centimeters are in 8 feet, given that 1 in. 2.54 cm.

Delicious77 [7]

Answer:

243.84

Step-by-step explanation:

We change feet to inches, and inches to cm.

1 ft = 12 in.

1 in. = 2.54 cm

8 ft * (12 in.)/(1 ft) * (2.54 cm)/(1 in.) = 243.84 cm

8 ft = 243.84 cm

3 0

3 years ago

Read 2 more answers

What is the logarithmic form of the solution to 102t = 9?

Harman [31]

Equivalent equations are equations that have the same value

The equation in logarithmic form is $t = \frac{\log(9)}{2}$

<h3>How to rewrite the equation</h3>

The expression is given as:

$10^{2t} = 9$

Take the logarithm of both sides

$\log(10^{2t}) = \log(9)$

Apply the power rule of logarithm

$2t\log(10) = \log(9)$

Divide both sides by log(10)

$2t = \frac{\log(9)}{\log(10)}$

Apply change of base rule

$2t = \log_{10}(9)$

Divide both sides by 2

$t = \frac{\log_{10}(9)}{2}$

Rewrite as:

$t = \frac{\log(9)}{2}$

Hence, the equation in logarithmic form is $t = \frac{\log(9)}{2}$

Read more about logarithms at:

brainly.com/question/25710806

8 0

3 years ago

Martin says that f(x)=2(4)^x starts at 4 and has a constant ratio of 2. What error did Martin make? Explain.

grigory [225]

Answer:

A general exponential equation is written as:

f(x) = A*(r)^x

Where:

A is the initial value, this is the value that the function takes when x = 0.

r is the rate of growth

x is the variable.

In this case, the function is:

f(x) = 2*(4)^x

The initial value is:

f(0) = 2*(4)^0 = 2*1 = 2

The initial value is 2.

And the rate of growth is 4.

So Martin seems to mixed these two values (he said initial value = 4, and rate = 2, which is the opposite of what we found)

8 0

3 years ago

Find the number of elements in A1 ∪A2 ∪A3 if there are 100 elements in A1, 1000 in A2, and 10,000 in A3 if a) A1 ⊆ A2 and A2 ⊆ A

Stella [2.4K]

Answer:

Step-by-step explanation:

Given that there are 3 sets such that there are 100 elements in A1, 1000 in A2, and 10,000 in A3

a) If A1 ⊆ A2 and A2 ⊆ A3

then union will contain the same number of elements as that of A3

i.e. $n(A1 ∪A2 ∪A3)=n(A3) =10000$

b) If the sets are pairwise disjoint.

union will contain the sum of elements of each set

$n(A1 ∪A2 ∪A3) = 100+1000+10000=11100$

c) If there are two elements common to each pair of sets and one element in all three sets

We subtract common elements pairwise and add common element in 3

i.e. $n(A1 ∪A2 ∪A3) = 100+1000+10000-2-2-2+1\\= 10995$

5 0

3 years ago

You need to invest $1000 in a bank account and are give two options. The first option is to earn $50 every month you leave the m

garri49 [273]

Answer:

<h3><u>Option 1</u></h3>

Earn $50 every month.

Let x = number of months the money is left in the account
Let y = the amount in the account
Initial amount = $1,000

$\implies y = 50x + 1000$

This is a <u>linear function</u>.

<h3><u>Option 2</u></h3>

Earn 3% interest each month.

(Assuming the interest earned each month is <u>compounding interest</u>.)

Let x = number of months the money is left in the account
Let y = the amount in the account
Initial amount = $1,000

$\implies y = 1000(1.03)^x$

This is an <u>exponential function</u>.

<h3><u>Table of values</u></h3>

<u />

$\large \begin{array}{| c | l | l |}\cline{1-3} & \multicolumn{2}{|c|}{\sf Account\:Balance} \\ \cline{1-3} & \sf Option\:1 & \sf Option\:2 \\\sf Month & \sf \$50\:per\:mth & \sf 3\%\:per\:mth \\\cline{1-3} 0 & \$1000 & \$1000 \\\cline{1-3} 1 & \$1050 & \$1030 \\\cline{1-3} 2 & \$1100 & \$1060.90 \\\cline{1-3} 3 & \$1150 & \$1092.73 \\\cline{1-3} 4 & \$1200 & \$1125.51 \\\cline{1-3} 5 & \$1250 & \$1159.27 \\\cline{1-3} 6 & \$1300 & \$1194.05 \\\cline{1-3} 7 & \$1350 & \$1229.87 \\\cline{1-3}\end{array}$

From the table of values, it appears that <u>Account Option 1</u> is the best choice, as the accumulative growth of this account is higher than the other account option.

However, there will be a point in time when Account Option 2 starts accruing more than Account Option 2 each month. To find this, graph the two functions and find the <u>point of intersection</u>.

From the attached graph, Account Option 1 accrues more until month 32. From month 33, Account Option 2 accrues more in the account.

<h3><u>Conclusion</u></h3>

If the money is going to be invested for less than 33 months then Account Option 1 is the better choice. However, if the money is going to be invested for 33 months or more, then Account Option 2 is the better choice.

3 0

2 years ago