Question 7 Write the integer that is equivalent to |10|

In 15 minutes Frankita created 15 drawings. In 30 minutes she created 17 drawings. In 45 minutes she created 20 drawings. In 60

SVETLANKA909090 [29]

Answer:

Frankita will create 29 drawings at the end of 75 minutes.

Step-by-step explanation:

Now below shown is the pattern used by Frankita:

15 min = 15 drawing

30 min = 17 drawing ⇒ increase by 2 (15+2 = 17)

45 min = 20 drawing ⇒ increase by 3 (17+3 = 20)

60 min = 24 drawing ⇒ increase by 4 (20+4 = 24)

75 min = 29 drawing ⇒ increase by 5 (24+5 = 29)

Hence We can say that Frankita will create 29 drawings at the end of 75 minutes.

8 0

4 years ago

One side of a rectangle is five times as long as the other side. If the perimeter is 72 meters, what is the length of the shorte

Yakvenalex [24]

6 meters

72 = x+x+5x+5x
72 = 12x
6 = x
the shorter side is 6 meters

3 0

4 years ago

Read 2 more answers

If x3= 27 and y3 =125, what is the value of y-x

Leno4ka [110]

Answer:

32 2/3

(32.6 repeating)

Step-by-step explanation:

29/3 is 9 (just make it not 3 Xs,) so x=9

125/3 is 41 2/3 so y= 41 2/3

41 2/3 - 9 = 32 2/3

5 0

3 years ago

Read 2 more answers

What is the opposite integer of 6

kotykmax [81]

It would be negative six.

3 0

3 years ago

Solve the given initial-value problem. x' = 1 2 0 1 − 1 2 x, x(0) = 2 7

Ilia_Sergeevich [38]

I'll go out on a limb and guess the system is

$\mathbf x'=\begin{bmatrix}\frac12&0\\1&-\frac12\end{bmatrix}\mathbf x$

with initial condition $\mathbf x(0)=\begin{bmatrix}2&7\end{bmatrix}^\top$ . The coefficient matrix has eigenvalues $\lambda$ such that

$\begin{vmatrix}\frac12-\lambda&0\\1&-\frac12-\lambda\end{vmatrix}=\lambda^2-\dfrac14=0\implies\lambda=\pm\dfrac12$

The corresponding eigenvectors $\eta$ are such that

$\lambda=\dfrac12\implies\begin{bmatrix}\frac12-\frac12&0\\1&-\frac12-\frac12\end{bmatrix}\eta=\begin{bmatrix}0&0\\1&-1\end{bmatrix}\eta=\begin{bmatrix}0\\0\end{bmatrix}$
$\implies\eta=\begin{bmatrix}1\\1\end{bmatrix}$

$\lambda=-\dfrac12\implies\begin{bmatrix}\frac12+\frac12&0\\1&-\frac12+\frac12\end{bmatrix}\eta=\begin{bmatrix}1&0\\1&0\end{bmatrix}\eta=\begin{bmatrix}0\\0\end{bmatrix}$
$\implies\eta=\begin{bmatrix}0\\1\end{bmatrix}$

So the characteristic solution to the ODE system is

$\mathbf x(t)=C_1\begin{bmatrix}1\\1\end{bmatrix}e^{t/2}+C_2\begin{bmatrix}0\\1\end{bmatrix}e^{-t/2}$

When $t=0$ , we have

$\begin{bmatrix}2\\7\end{bmatrix}=C_1\begin{bmatrix}1\\1\end{bmatrix}+C_2\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}C_1\\C_1+C_2\end{bmatrix}$

from which it follows that $C_1=2$ and $C_2=5$ , making the particular solution to the IVP

$\mathbf x(t)=2\begin{bmatrix}1\\1\end{bmatrix}e^{t/2}+5\begin{bmatrix}0\\1\end{bmatrix}e^{-t/2}$

$\mathbf x(t)=\begin{bmatrix}2e^{t/2}\\2e^{t/2}+5e^{-t/2}\end{bmatrix}$

5 0

4 years ago