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Naily [24]
3 years ago
15

What is the actual length?

Mathematics
1 answer:
Andre45 [30]3 years ago
6 0

Answer:

Step-by-step explanation:

2ft

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Lars deposited $50 into a savings account for which interest is compounded quarterly. According to the rule of 72, what interest
Ivanshal [37]

Answer: C 2.5%


Step-by-step explanation:

The "Rule of 72" is a easy way to calculate how much time an investment will take to double with a given fixed annual rate of interest.

Just we have to divide 72 by the annual rate of return(r), we can get a rough estimate of how many years it will take to double the initial investment .

Now, in given problem: Let 'r' be the rate of interest

Time to double the amount=29 years

Thus by rule 72 ,

\frac{72}{r}=29\\\Rightarrow\ r=\frac{72}{29}=2.4827\%\approx2.5\%

Therefore, C is the right option.


3 0
3 years ago
Read 2 more answers
I cant figure this out its x+1 &gt; 8 then<br> x= __. its inequalities math thank you
olganol [36]

Answer:X is any number more than 7

Source:trust me bro

4 0
3 years ago
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Which shape must have opposite sides that are parallel and congruent, and diagonals that are perpendicular bisectors of each oth
n200080 [17]

Answer:

Rectangle or square

Step-by-step explanation:

The diagonals are perpendicular bisectors of each other. C) All four angles of the parallelogram are right angles. Congruent diagonals eliminates the rhombus. This leaves the rectangle or the square.

8 0
3 years ago
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If f(–5) = 0, what are all the factors of the function f (x ) = x cubed minus 19 x + 30? Use the Remainder Theorem.
oksian1 [2.3K]
(x + 5) (x - 2) (x - 3)

4 0
2 years ago
Evaluate the integral. (sec2(t) i t(t2 1)8 j t7 ln(t) k) dt
polet [3.4K]

If you're just integrating a vector-valued function, you just integrate each component:

\displaystyle\int(\sec^2t\,\hat\imath+t(t^2-1)^8\,\hat\jmath+t^7\ln t\,\hat k)\,\mathrm dt

=\displaystyle\left(\int\sec^2t\,\mathrm dt\right)\hat\imath+\left(\int t(t^2-1)^8\,\mathrm dt\right)\hat\jmath+\left(\int t^7\ln t\,\mathrm dt\right)\hat k

The first integral is trivial since (\tan t)'=\sec^2t.

The second can be done by substituting u=t^2-1:

u=t^2-1\implies\mathrm du=2t\,\mathrm dt\implies\displaystyle\frac12\int u^8\,\mathrm du=\frac1{18}(t^2-1)^9+C

The third can be found by integrating by parts:

u=\ln t\implies\mathrm du=\dfrac{\mathrm dt}t

\mathrm dv=t^7\,\mathrm dt\implies v=\dfrac18t^8

\displaystyle\int t^7\ln t\,\mathrm dt=\frac18t^8\ln t-\frac18\int t^7\,\mathrm dt=\frac18t^8\ln t-\frac1{64}t^8+C

8 0
3 years ago
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