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3 years ago

Solve the inequality for u. -10 ≤ - 5u Simplify your answer as much as possible.

Mathematics

1 answer:

inn [45]3 years ago

4 0

Answer:

u ≤ 2

Step-by-step explanation:

Given

- 10 ≤ - 5u

Divide both sides by - 5 , reversing the symbol as a result of dividing by a negative quantity.

2 ≥ u , then

u ≤ 2

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Find the 10th term 3,15,75,375

vova2212 [387]

Answer:

5859375

Step-by-step explanation:

These are the terms of a geometric sequence with n th term

$a_{n}$ = $a_{1}$ $r^{n-1}$

where r is the common ratio and $a_{1}$ the first term

here r = $\frac{375}{75}$ = $\frac{75}{15}$ = $\frac{15}{3}$ = 5 and $a_{1}$ = 3, hence

$a_{10}$ = 3 × $(5)^{9}$ = 5859375

5 0

3 years ago

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

Test each statement below to see if it is reversible. If so, write it as a true biconditional. If not, write not reversible

FinnZ [79.3K]

Statement: Complementary angles are two angles with measures that have a sum of 90.
Reverse: Two angles with measures that have a sum of 90 are complementary angles.

Here the reverse is true, therefore, the statement is true biconditional.

Statement: A rectangle is a four-sided figure with at least one right angle.
Reverse: A four sided figure with at least one right angle is a rectangle.

Here, the reverse is not true, therefore, the statement is not reversible.

3 0

3 years ago

Read 2 more answers

If two lines are parallel they also perpendicular?

egoroff_w [7]

Answer:

Step-by-step explanation:

Parallel lines do not intersect. Perpendicular lines intersect at right angles. Lines cannot be both.

3 0

3 years ago

Read 2 more answers

Do number one please and make sure to show your work thank you