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posledela
3 years ago
9

Find all real values of t that satisfy the equation t^2 = 324. If you find more than one, then list your values in increasing or

der, separated by commas.
Mathematics
1 answer:
vredina [299]3 years ago
4 0

Answer:

t = -18, 18

Step-by-step explanation:

Take the square root of both sides of the equation.

... t = ±√324 = ±18

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) Which expressions are equivalent to 45? Choose ALL that apply.
VMariaS [17]
4.5 and maybe 16-64. If they count -
5 0
2 years ago
Evaluate the function:<br><br> f(x)=4x−1, find f(9)
Eddi Din [679]

Answer:

f(9) = 35

Step-by-step explanation:

To solve this, you can take 9 and replace both x's with it and then complete the equation;

f(x) = 4x - 1

f(9) = 4(9) - 1

f(9) = 36 - 1

f(9) = 35

Hope this helps!

3 0
2 years ago
7 qt. of a 27% saline solution was mixed
mamaluj [8]

Answer:

24% saline

Step-by-step explanation:

Find how much saline is in both solutions:

7(0.27)

= 1.89

3(0.17)

= 0.51

Add this together:

1.89 + 0.51

= 2.4

Since there are 10 quarts in total, the concentration of the mixture can be found by dividing 2.4 by 10:

2.4/10

= 0.24

So, the concentration of the mixture is 24% saline

4 0
3 years ago
$210 at 8% for 7 years
Genrish500 [490]
The answer is $327.60
7 0
3 years ago
Consider the differential equation:
Wewaii [24]

(a) Take the Laplace transform of both sides:

2y''(t)+ty'(t)-2y(t)=14

\implies 2(s^2Y(s)-sy(0)-y'(0))-(Y(s)+sY'(s))-2Y(s)=\dfrac{14}s

where the transform of ty'(t) comes from

L[ty'(t)]=-(L[y'(t)])'=-(sY(s)-y(0))'=-Y(s)-sY'(s)

This yields the linear ODE,

-sY'(s)+(2s^2-3)Y(s)=\dfrac{14}s

Divides both sides by -s:

Y'(s)+\dfrac{3-2s^2}sY(s)=-\dfrac{14}{s^2}

Find the integrating factor:

\displaystyle\int\frac{3-2s^2}s\,\mathrm ds=3\ln|s|-s^2+C

Multiply both sides of the ODE by e^{3\ln|s|-s^2}=s^3e^{-s^2}:

s^3e^{-s^2}Y'(s)+(3s^2-2s^4)e^{-s^2}Y(s)=-14se^{-s^2}

The left side condenses into the derivative of a product:

\left(s^3e^{-s^2}Y(s)\right)'=-14se^{-s^2}

Integrate both sides and solve for Y(s):

s^3e^{-s^2}Y(s)=7e^{-s^2}+C

Y(s)=\dfrac{7+Ce^{s^2}}{s^3}

(b) Taking the inverse transform of both sides gives

y(t)=\dfrac{7t^2}2+C\,L^{-1}\left[\dfrac{e^{s^2}}{s^3}\right]

I don't know whether the remaining inverse transform can be resolved, but using the principle of superposition, we know that \frac{7t^2}2 is one solution to the original ODE.

y(t)=\dfrac{7t^2}2\implies y'(t)=7t\implies y''(t)=7

Substitute these into the ODE to see everything checks out:

2\cdot7+t\cdot7t-2\cdot\dfrac{7t^2}2=14

5 0
3 years ago
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