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dlinn [17]
2 years ago
6

The volume (V) of a rectangular prism whose height is 12 cm and base is a square with side lengths s cm: V = 12s to the second p

ower
Mathematics
1 answer:
lions [1.4K]2 years ago
4 0

Answer: 12s^2

you're welcome :D

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What is the value of x?​
guajiro [1.7K]

Answer:

x ≈ 8.66

Step-by-step explanation:

Using the sine ratio in the right triangle

sin60° = \frac{opposite}{hypotenuse} = \frac{x}{10}

Multiply both sides by 10

10 × sin60° = x, thus

x ≈ 8.66 ( to 2 dec. places )

4 0
3 years ago
Eli invested $ 330 $330 in an account in the year 1999, and the value has been growing exponentially at a constant rate. The val
Cloud [144]

Answer:

The value of the account in the year 2009 will be $682.

Step-by-step explanation:

The acount's balance, in t years after 1999, can be modeled by the following equation.

A(t) = Pe^{rt}

In which A(t) is the amount after t years, P is the initial money deposited, and r is the rate of interest.

$330 in an account in the year 1999

This means that P = 330

$590 in the year 2007

2007 is 8 years after 1999, so P(8) = 590.

We use this to find r.

A(t) = Pe^{rt}

590 = 330e^{8r}

e^{8r} = \frac{590}{330}

e^{8r} = 1.79

Applying ln to both sides:

\ln{e^{8r}} = \ln{1.79}

8r = \ln{1.79}

r = \frac{\ln{1.79}}{8}

r = 0.0726

Determine the value of the account, to the nearest dollar, in the year 2009.

2009 is 10 years after 1999, so this is A(10).

A(t) = 330e^{0.0726t}

A(10) = 330e^{0.0726*10} = 682

The value of the account in the year 2009 will be $682.

4 0
3 years ago
ANSWER THIS ASAP DONT SEND A FILE AT ALL !! I NEED THE ANSWER TO QUESTION
jasenka [17]

Answer: just use phot math  

Step-by-step explanation:

4 0
2 years ago
Find the oth term of the geometric sequence 7, 14, 28, ...
yaroslaw [1]

Answer:

The nth term of the geometric sequence 7, 14, 28, ... is:

a_n=7\cdot \:2^{n-1}

Step-by-step explanation:

Given the geometric sequence

7, 14, 28, ...

We know that a geometric sequence has a constant ratio 'r' and is defined by

a_n=a_1\cdot r^{n-1}

where a₁ is the first term and r is the common ratio

Computing the ratios of all the adjacent terms

\frac{14}{7}=2,\:\quad \frac{28}{14}=2

The ratio of all the adjacent terms is the same and equal to

r=2

now substituting r = 2 and a₁ = 7 in the nth term

a_n=a_1\cdot r^{n-1}

a_n=7\cdot \:2^{n-1}

Therefore, the nth term of the geometric sequence 7, 14, 28, ... is:

a_n=7\cdot \:2^{n-1}

6 0
2 years ago
What is the sum of 1/4 and 3/8
kupik [55]
\dfrac{1}{4}+\dfrac{3}{8}=(*)\\Find\ LCD\ (Least\ Common\ Denominator)\\list\ the\ multiples\\of\ 4:\ \ 0;\ 4;\ \fbox8;\ 12;...\\of\ 8:\ \ 0;\ \fbox8;\ 16;\ 24;\ ...\\LCD\left(\dfrac{1}{4};\ \dfrac{3}{8}\right)=8\\\\(*)=\dfrac{1\cdot2}{4\cdot2}+\dfrac{3}{8}=\dfrac{2}{8}+\dfrac{3}{8}=\dfrac{2+3}{8}=\dfrac{5}{8}
4 0
3 years ago
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