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emmainna [20.7K]
3 years ago
10

I need help with number 16

Mathematics
1 answer:
Alex3 years ago
7 0

Answer:72 in

Step-by-step explanation:

When we are looking for the area of something we always multiply the numbers they give us.

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After 6 years, what is the total amount of a compound interest investment of $35,000 at 4% interest, compounded quarterly?
sammy [17]

Answer:

<h2>$9441</h2>

Step-by-step explanation:

The interest will be compounded quarterly every year, it means in each year the interest will be calculated 4 times.

In 6 years in total 6 \times4 = 24 times the interest will be calculated.

The yearly interest rate is 4%. Hence, the quaterly interest rate will be \frac{4}{4} = 1%.

Hence, after calculating 24 times, the amount will be turned to 35000 \times (\frac{101}{100} )^{24} = 44440.7127≅  44441.

Hence, the total compound interest is $(44441 - 35000) = $9441

4 0
3 years ago
Read 2 more answers
Edin has £300 in his saving account.
Veronika [31]

Answer:

$45

Step-by-step explanation:

300/100 = $3 = 1%

$3 x 5 = $15 = 5%

$15 x 3 years = $45

4 0
3 years ago
Read 2 more answers
100 POINTS!!!!
Vladimir [108]

<em>There is no specific requirement in the question, but I'm assuming you need to compute the time needed for Alexis reach 1,000,000 Instagram followers</em>

Answer:

t= 462.82\ days

Step-by-step explanation:

<u>Exponential Growth </u>

When the number of observed elements grows as the previous value multiplied by a constant ratio, we have exponential growth. The formula to model such situations is

\displaystyle f(x) = f_o(1 + r)^t

Where f_o is the initial value of f, 1 + r is the constant ratio, and t is the time expressed in half days (12 hours)

The initial value is 100 Instant followers, thus:

\displaystyle f(x) = 100(1.01)^t

We need to know when the number of followers will reach 1,000,000. Setting up the equation

\displaystyle 100(1.01)^t=1,000,000

Simplifying by 100

\displaystyle (1.01)^t=10,000

Taking logarithms

\displaystyle t\ log(1.01)=log\ 10,000

\displaystyle t\ log(1.01)=4

Solving for t

\displaystyle t=\frac{4}{log(1.01)}=925.63 periods of 12 hrs

t= 462.82\ days

7 0
4 years ago
small cubes with edge lengths of 1/4 inch will be packed into the right rectangular prism shown.( the base is 4 1/2, the width i
ss7ja [257]

General Idea:

We need to find the volume of the small cube given the side length of the small cube as 1/4 inch.

Also we need to find the volume of the right rectangular prism with the given dimension (the height is 4 1/2, the width is 5, and the length is 3 3/4).

To find the number of small cubes that are needed to completely fill the right rectangular prism, we need to divide volume of right rectangular prism by volume of each small cube.

Formula Used:

Volume \; of \; Cube = a^3 \; \\\{where \; a \; is \; side \; length \; of \; cube\}\\\\Volume \; of  \; Right \; Rectangular  \; Prism=L \times W \times H\\\{Where  \; L \; is \; Length, \; W \; is \; Width, \;and  \; H \; is \; Height\}

Applying the concept:

Volume of Small Cube:

V_{cube}= (\frac{1}{4}  )^3= \frac{1}{64} \; in^3\\\\V_{Prism}=  3 \frac{3}{4}  \times 5 \times  4 \frac{1}{2}  = \frac{15}{4}  \times \frac{5}{1}  \times \frac{9}{2}  = \frac{675}{8}  \\\\Number \; of \; small \; cubes= \frac{V_{Prism}}{V_{Cube}}   = \frac{675}{8}  \div \frac{1}{64}  \\\\Flip \; the \; second \; fraction\; and \; multiply \; with \; the \; first \; fraction\\\\Number \; of \; small \; cubes \;= \frac{675}{8} \times \frac{64}{1}   = 5400

Conclusion:

The number of small cubes with side length as 1/4 inches that are needed to completely fill the right rectangular prism whose height is 4 1/2 inches, width is 5 inches, and length is 3 3/4 inches is <em><u>5400 </u></em>

4 0
3 years ago
Read 2 more answers
Somebody help me pleaseeeeeeeeee
Fudgin [204]

Answer:

Write the inequality: 2.90h + 4.75

Solve the inequality: 2.90h + 4.75 = 39.55

Step-by-step explanation:

I’m not really sure if its correct I got it off somebody else’s brainly account

I’m actually currently doing the test right now and ive been using you’re page for the answers, thanks you’re account was a big help .

6 0
3 years ago
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