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BaLLatris [955]

3 years ago

6

In a bowl, there are gum balls. There are 5 red, 7 blue, 3 yellow and 11 purple.

1 answer:

ipn [44]3 years ago

6 0

Answer:

10 is A

11 is D

12 is C

and 13 is B

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Find the 5th term of the sequence which has a first of 1 and a common ratio of 2.3

sveticcg [70]

Hello :

the nth term is : An =A1 × r^(n-1)

A1 = 1 r = 2.3 n = 5

A5 =1 × (2.3)^(5-1)=27.9841

3 0

3 years ago

Please help me, very important 20 points!

kipiarov [429]

x=96

all interior angles equal 180

Subtract exterior andgles by 180 to get interior angles because they are supplementary

180-134=46

180-130=50

Get the third angle inside the triangle

180-50-46=84

subtract by 180 to get x because they are supplementary

180-84=96

Therefore x =96

4 0

3 years ago

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Rashid [163]

Answer:

D, <VMY cannot be a right angle, its an obtuse angle (more specifically, ab a 120 degree angle)

6 0

3 years ago

Consider the initial value problem y′+5y=⎧⎩⎨⎪⎪0110 if 0≤t<3 if 3≤t<5 if 5≤t<[infinity],y(0)=4. y′+5y={0 if 0≤t<311 i

rosijanka [135]

It looks like the ODE is

$y'+5y=\begin{cases}0&\text{for }0\le t$

with the initial condition of $y(0)=4$ .

Rewrite the right side in terms of the unit step function,

$u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t$

In this case, we have

$\begin{cases}0&\text{for }0\le t$

The Laplace transform of the step function is easy to compute:

$\displaystyle\int_0^\infty u(t-c)e^{-st}\,\mathrm dt=\int_c^\infty e^{-st}\,\mathrm dt=\frac{e^{-cs}}s$

So, taking the Laplace transform of both sides of the ODE, we get

$sY(s)-y(0)+5Y(s)=\dfrac{e^{-3s}-e^{-5s}}s$

Solve for $Y(s)$ :

$(s+5)Y(s)-4=\dfrac{e^{-3s}-e^{-5s}}s\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}{s(s+5)}+\dfrac4{s+5}$

We can split the first term into partial fractions:

$\dfrac1{s(s+5)}=\dfrac as+\dfrac b{s+5}\implies1=a(s+5)+bs$

If $s=0$ , then $1=5a\implies a=\frac15$ .

If $s=-5$ , then $1=-5b\implies b=-\frac15$ .

$\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}5\left(\frac1s-\frac1{s+5}\right)+\dfrac4{s+5}$

$\implies Y(s)=\dfrac15\left(\dfrac{e^{-3s}}s-\dfrac{e^{-3s}}{s+5}-\dfrac{e^{-5s}}s+\dfrac{e^{-5s}}{s+5}\right)+\dfrac4{s+5}$

Take the inverse transform of both sides, recalling that

$Y(s)=e^{-cs}F(s)\implies y(t)=u(t-c)f(t-c)$

where $F(s)$ is the Laplace transform of the function $f(t)$ . We have

$F(s)=\dfrac1s\implies f(t)=1$

$F(s)=\dfrac1{s+5}\implies f(t)=e^{-5t}$

We then end up with

$y(t)=\dfrac{u(t-3)(1-e^{-5t})-u(t-5)(1-e^{-5t})}5+5e^{-5t}$

3 0

3 years ago

1. S = 10 mm<br> V= S×S×S<br> =___×___×___<br> =____ mm3<br><br>

Minchanka [31]

Hi there!

»»————-　★　————-««

I believe your answer is:

$V=1000\text{mm}^3$

»»————-　★　————-««

Here’s why:

⸻⸻⸻⸻

I am assuming by the infomation given that the figure is a cube.

⸻⸻⸻⸻

$\boxed{\text{Finding the volume of the cube...}}\\\\S = 10mm; V= s^3\\--------------\\\rightarrow V = 10^3\\\\\rightarrow V = 10 * 10 * 10\\\\\rightarrow \boxed{V=1000\text{mm}^3}$

⸻⸻⸻⸻

»»————-　★　————-««

Hope this helps you. I apologize if it’s incorrect.

3 0

3 years ago