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3 years ago

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The temperatures recorded at three different times on a november day in fairbanks, alaska were -2 -5 -12. Which inequality corre

ctly compares thease temperatures

1 answer:

Gelneren [198K]3 years ago

3 0

Answer:

$-12 < -5 < -2$

Step-by-step explanation:

Given

$Temperature = \{-2, -5,-12\}$

Required

Represent as an inequality

On a number line:

-12 comes before -2 and -5

-5 comes before -2

This means that:

-12 is smaller than -5 and -5 is smaller than -2.

So, we have:

$-12 < -5$ and $-5 < -2$

Combine both inequalities

$-12 < -5 < -2$

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Use the method of undetermined coefficients to solve the given nonhomogeneous system. x' = −1 5 −1 1 x + sin(t) −2 cos(t)

AlekseyPX

It looks like the system is

$x' = \begin{bmatrix} -1 & 5 \\ -1 & 1 \end{bmatrix} x + \begin{bmatrix} \sin(t) \\ -2 \cos(t) \end{bmatrix}$

Compute the eigenvalues of the coefficient matrix.

$\begin{vmatrix} -1 - \lambda & 5 \\ -1 & 1 - \lambda \end{vmatrix} = \lambda^2 + 4 = 0 \implies \lambda = \pm2i$

For $\lambda = 2i$ , the corresponding eigenvector is $\eta=\begin{bmatrix}\eta_1&\eta_2\end{bmatrix}^\top$ such that

$\begin{bmatrix} -1 - 2i & 5 \\ -1 & 1 - 2i \end{bmatrix} \begin{bmatrix} \eta_1 \\ \eta_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

Notice that the first row is 1 + 2i times the second row, so

$(1+2i) \eta_1 - 5\eta_2 = 0$

Let $\eta_1 = 1-2i$ ; then $\eta_2=1$ , so that

$\begin{bmatrix} -1 & 5 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 - 2i \\ 1 \end{bmatrix} = 2i \begin{bmatrix} 1 - 2i \\ 1 \end{bmatrix}$

The eigenvector corresponding to $\lambda=-2i$ is the complex conjugate of $\eta$ .

So, the characteristic solution to the homogeneous system is

$x = C_1 e^{2it} \begin{bmatrix} 1 - 2i \\ 1 \end{bmatrix} + C_2 e^{-2it} \begin{bmatrix} 1 + 2i \\ 1 \end{bmatrix}$

The characteristic solution contains $\cos(2t)$ and $\sin(2t)$ , both of which are linearly independent to $\cos(t)$ and $\sin(t)$ . So for the nonhomogeneous part, we consider the ansatz particular solution

$x = \cos(t) \begin{bmatrix} a \\ b \end{bmatrix} + \sin(t) \begin{bmatrix} c \\ d \end{bmatrix}$

Differentiating this and substituting into the ODE system gives

$-\sin(t) \begin{bmatrix} a \\ b \end{bmatrix} + \cos(t) \begin{bmatrix} c \\ d \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ -1 & 1 \end{bmatrix} \left(\cos(t) \begin{bmatrix} a \\ b \end{bmatrix} + \sin(t) \begin{bmatrix} c \\ d \end{bmatrix}\right) + \begin{bmatrix} \sin(t) \\ -2 \cos(t) \end{bmatrix}$

$\implies \begin{cases}a - 5c + d = 1 \\ b - c + d = 0 \\ 5a - b + c = 0 \\ a - b + d = -2 \end{cases} \implies a=\dfrac{11}{41}, b=\dfrac{38}{41}, c=-\dfrac{17}{41}, d=-\dfrac{55}{41}$

Then the general solution to the system is

$x = C_1 e^{2it} \begin{bmatrix} 1 - 2i \\ 1 \end{bmatrix} + C_2 e^{-2it} \begin{bmatrix} 1 + 2i \\ 1 \end{bmatrix} + \dfrac1{41} \cos(t) \begin{bmatrix} 11 \\ 38 \end{bmatrix} - \dfrac1{41} \sin(t) \begin{bmatrix} 17 \\ 55 \end{bmatrix}$

7 0

2 years ago

10) Data Distribution

murzikaleks [220]

Answer:

mean 1.5

mode a 0

median 1

Step-by-step explanation:

4 0

3 years ago

If LM = 6, what is the perimeter of △PKQ?

Tanya [424]

16 i believe
ur welcome

7 0

3 years ago

Read 2 more answers

I need help with question please help me

larisa [96]

Answer:

189.3 unit^2.

Step-by-step explanation:

The total area = area of the rectangle + the area of the semicircle

= 15*10 + 0.5 π*5^2 ( note the radius = 5)

= 150 + 12.5 π

= 150 + 39.3

= 189.3 unit^2.

7 0

3 years ago

Given the endpoints D(-2, 3) and L(4,1)<br> Find the midpoint of DL.

Igoryamba

Answer:

Midpoint = (1,2)

Step-by-step explanation:

Using coordinates: D(-2,3) and L(4,1)

Using the midpoint formula:

Midpoint of DL = (D1 + L1) / 2, (D2 + L2) / 2

Midpoint of DL = (-2 + 4) / 2, (3 + 1) / 2

Midpoint of DL = (2) / 2, (4) / 2

Therefore the midpoint of DL is (1,2)

8 0

3 years ago