Explain how to estimate the quotient using compatible numbers. 27 and two-thirds divided by 3 and StartFraction 9 over 10 EndFra

Answer:

m = - 3

Step-by-step explanation:

a³ + 27 can be recognized as a sum of cubes, which factors generally as

a³ + b³ = (a + b)(a² - ab + b²). Therefore:

a³ + 27

= a³ + 3³

= (a + 3)(a² - 3a + 9).

By comparing a² - 3a + 9 to a² + ma + 9, we find that

m = - 3.

To determine the values of b that fulfill 3(2b+3)^2 = 36

we start with
3(2b+3)^2 = 36
Divide both sides by 3
(2b+3)^2 = 12
Next, take the square root of both sides
(2b+3)} = (+ /-) \sqrt{12} \\ 2b=(+ /-) \sqrt{12}-3

b1=\frac{\sqrt{12}}{2} -\frac{3}{2}
b1=\sqrt{3} -\frac{3}{2}

b2=\frac{-\sqrt{12}}{2} -\frac{3}{2}
b2=-\sqrt{3} -\frac{3}{2}
Thus,

the solutions for b are
b1=\sqrt{3} -\frac{3}{2}
b2=-\sqrt{3} -\frac{3}{2}

Utilizing the normal distribution and the central limit theorem, there's a 0.0284 or 2.84% chance of observing a sample mean mass of 695g or less.

The question clearly seeks the highest values from both functions, meaning the vertices of each.

<span>The graph depicting the path of Ed’s football indicates the vertex's coordinates (the peak of the graph).
</span>
Specifically,

(h,k) = (1.5, 7.5)

Where (h,k) represents the vertex's location.

Conversely,<span>the trajectory of Steve's football is defined by the equation:
y = - 2x
²</span> + 5x + 4<span>To find the axis of symmetry, we use the formula:x = - b
÷ 2a
Where:

a = -2</span>
b = 5

Consequently,

x = - 5 ÷ - 4

x = 5 / 4




x = 1.25
Now substituting this x-value back into the main equation to determine y.
y = - 2x² + 5x + 4
y = - 2(1.25)² + 5(1.25) + 4

y = - 3.125 + 6.25 + 4


y = 7.125


Thus, the vertex (h,k) = (1.25, 7.125)
As observed from the calculations
Ed’s

<span>football attains a higher height.</span>

Answer:

u(t)  = -(3 + w^2 ) cos t /(1- w^2)cos t + 7 sin t + 8 cos wt /(1- w^2)

Step-by-step explanation:

The characteristic equation is k² + 1 = 0, which leads to k² = -1, resulting in k = ±i.

The roots are k = i or -i.

The general solution takes the form  u(x)=C₁cosx+C₂sinx.

Applying the method of undetermined coefficients, we have

Uc(t) = Pcos wt  + Qsin wt

Calculating the derivatives gives us Uc’(t) = -Pwsin wt  + Qwcos wt

And differentiating again yields Uc’’(t) = -Pw^2cos wt  - Qw^2sin wt

With the equation U’’ + u = 8cos wt, we substitute:

-Pw^2cos wt  - Qw^2sin wt + Pcos wt  + Qsin wt = 8cos wt.

This simplifies to (-Pw^2 + P) cos wt   + (-Qw^2 + Q) sin wt = 8cos wt.

From -Pw^2 + P = 8, we find P= 8  /(1- w^2).

From -Qw^2 + Q = 8, we can conclude Q = 0.

Thus, Uc(t) = Pcos wt  + Qsin wt = 8 cos wt /(1- w^2).

Combining gives us U(t) = uh(t ) + Uc(t)

     = C1cos t + c2 sin t + 8 cos wt /(1- w^2).

Initial conditions yield:

U(0) = C1cos(0) + c2 sin (0) + 8 cos (0) /(1- w^2)

Which leads us to C1 + 8 /(1- w^2) = 5

So C1 = 5 - 8 /(1- w^2) = -(3 + w^2 ) /(1- w^2).

Next, taking the derivative:

U’(t) = -C1 sin t + c2 cos t - 8 w sin wt /(1- w^2).

Evaluating at t = 0 gives us:

U’(0) = -C1 sin (0) + c2 cos (0) - 8 w sin (0) /(1- w^2) = 7.

Thus, c2 = 7.

  u(t)  = -(3 + w^2 ) cos t /(1- w^2)cos t + 7 sin t + 8 cos wt /(1- w^2)