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2 years ago

6

Enter the correct answer in the box. What are the solutions of this quadratic equation? X^2+164=16x Substitute the values of a a

nd b to complete the solutions.

1 answer:

saw5 [17]2 years ago

4 0

Answer:

x = 8 - 10i or 8 + 10i

Step-by-step explanation:

Our equation is given as:

x² + 164 = 16x

x² - 16x + 164 = 0

We solve using Almighty formula

x = -b ± √b² - 4ac/2a

Where : ax² + bx + c

For our Equation

a = 1, b = -16, c = 164

Hence,

x = -(-16) ± √-16² - 4 × 1 × 164/2 × 1

x = 16 ± √256 - 656/2

x = 16 ± √-400/2

Finding the roots

x = 16/2 ± √-400/2

x = 8 ± 20i/2

x = 8 ± 10i

Therefore, the value of x

= 8 - 10i or 8 + 10i

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The area of another rectangle is x² + 5x + 6. If one side measures (x + 2), what is the measure of the other

tiny-mole [99]

L=x+2
We need B

$\\ \sf\longmapsto LB=Area$

$\\ \sf\longmapsto B(x+2)=x^2+5x+6$

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8 0

2 years ago

X + 78 = 90 can someone solve this for me please?

kondaur [170]

Let's solve your equation step-by-step.

x+78=90

Step 1: Subtract 78 from both sides.

x+78−78=90−78

x=12

Please put brainliest!

8 0

2 years ago

Please help thank you.

ki77a [65]

Hello!

Let O be the center of the sphere, A the tangency point and B the location of the satellite.

m<ABO = 1/2m(b) = 1/2 * 138 = 69

m<OAB = 90

m<AOB = 180 - 90 - 69 = 21

21 * 2 = 42

The answer is 42°

Hope this helps!

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

2 years ago

ratelena [41]

Answer:

Positive and negative intervals on a graph

Step-by-step explanation:

The positive regions of a function are those intervals where the function is above the x-axis. It is where the y-values are positive (not zero). The negative regions of a function are those intervals where the function is below the x-axis. It is where the y-values are negative (not zero). hope this helps you :)

3 0

2 years ago

Read 2 more answers