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lesya [120]
3 years ago
11

Substitute the given values into the given formula and solve for the unknown variable. If necessary, round to one decimal place

Mathematics
1 answer:
ratelena [41]3 years ago
5 0

Answer:

T = Time = 3 years

Step-by-step explanation:

1 = PRT;

1 = Simple Interest = 5040

P = Principal = 24,000

R = Rate = 0.07

T = Time (years) = ? Unknown variable

Making T subject of the formula

T = 1/PR

T = 5040/24,000 × 0.07

T = 5040/1680

T = 3 years

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Answer:

-42

Step-by-step explanation:

1. 4(-2)

2. 3(-8)

3. -8 + -10 + -24

4. cobine like terms

5. -42

7 0
2 years ago
Consider the function f given by f(x)=x*(e^(-x^2)) for all real numbers x.
NISA [10]

Answer:

\frac{\sqrt{\pi}}{4}

Step-by-step explanation:

You are going to integrate the following function:

g(x)=x*f(x)=x*xe^{-x^2}=x^2e^{-x^2}  (1)

furthermore, you know that:

\int_0^{\infty}e^{-x^2}=\frac{\sqrt{\pi}}{2}

lets call to this integral, the integral Io.

for a general form of I you have In:

I_n=\int_0^{\infty}x^ne^{-ax^2}dx

furthermore you use the fact that:

I_n=-\frac{\partial I_{n-2}}{\partial a}

by using this last expression in an iterative way you obtain the following:

\int_0^{\infty}x^{2s}e^{-ax^2}dx=\frac{(2s-1)!!}{2^{s+1}a^s}\sqrt{\frac{\pi}{a}} (2)

with n=2s a even number

for s=1 you have n=2, that is, the function g(x). By using the equation (2) (with a = 1) you finally obtain:

\int_0^{\infty}x^2e^{-x^2}dx=\frac{(2(1)-1)!}{2^{1+1}(1^1)}\sqrt{\pi}=\frac{\sqrt{\pi}}{4}

5 0
3 years ago
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Answer:

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Step-by-step explanation:

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Hope it will help :)

7 0
2 years ago
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Step-by-step explanation:

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2 years ago
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igomit [66]

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16

Step-by-step explanation:

9 divided by 144 gives you 16

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2 years ago
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