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iogann1982 [59]

3 years ago

14

POINT MARKET COME GET YO POINTS

2 answers:

Vinvika [58]3 years ago

8 0

Answer:

ball are going to bw inserted into

Step-by-step explanation:

yo jaws

vfiekz [6]3 years ago

3 0

Answer:

Hi ansndvgsgshhshsjajaja

Step-by-step explanation:

I just want points

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Please Help!, Brainliest to whoever helps

Tanya [424]

B for sure!!!!!!!!!!!!!!!!!!!

6 0

2 years ago

plz help me and I kindly ask that u show work if possible? Plz don't answer if u don't know how to do the problem. PLZ help

soldier1979 [14.2K]

part A is just simple multiplication
since there are 2 6 sided number cubes there are 36 possible outcomes(6 sides *6 sides)

Part B is a little trickier. First off, you must find how to get seven.
1. 1+6 and 6+1

2. 2+5 and 5+2

3.3+4 and 4+3

therefore there are six combinations

you put that number over the total combos and get 6/36 or 1/6.

Hope this helps!

6 0

2 years ago

Read 2 more answers

What is the inverse of f(x)=(x+6)2 for x≥–6 where function g is the inverse of function f? g(x)=x√−6, x≥0 g(x)=x−6−−−−√, x≥6 g(x

avanturin [10]

Answer:

$g(x)=\sqrt{x}-6$ , x>=0

Step-by-step explanation:

$f(x)=(x+6)^2$

To find the inverse function , replace f(x) by y

$y=(x+6)^2$

Replace x with y and y with x

$x=(y+6)^2$

Solve the equation for y

take square root on both sides

$+\sqrt{x} =y+6$

Now subtract 6 from both sides

$+\sqrt{x}-6=y$

replace y with g(x)

$g(x)=\sqrt{x}-6$

8 0

3 years ago

There are 18 students in a music class , 6 of them are male. What is the probability of choosing a male student from the class?

Crank

Answer:

0.33 repeating

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

2 points) Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One

Stells [14]

$y'=(t+y)^2-1$

Substitute $u=t+y$ , so that $u'=y'$ , and

$u'=u^2-1$

which is separable as

$\dfrac{u'}{u^2-1}=1$

Integrate both sides with respect to $t$ . For the integral on the left, first split into partial fractions:

$\dfrac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)=1$

$\displaystyle\int\frac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)\,\mathrm dt=\int\mathrm dt$

$\dfrac12(\ln|u-1|-\ln|u+1|)=t+C$

Solve for $u$ :

$\dfrac12\ln\left|\dfrac{u-1}{u+1}\right|=t+C$

$\ln\left|1-\dfrac2{u+1}\right|=2t+C$

$1-\dfrac2{u+1}=e^{2t+C}=Ce^{2t}$

$\dfrac2{u+1}=1-Ce^{2t}$

$\dfrac{u+1}2=\dfrac1{1-Ce^{2t}}$

$u=\dfrac2{1-Ce^{2t}}-1$

Replace $u$ and solve for $y$ :

$t+y=\dfrac2{1-Ce^{2t}}-1$

$y=\dfrac2{1-Ce^{2t}}-1-t$

Now use the given initial condition to solve for $C$ :

$y(3)=4\implies4=\dfrac2{1-Ce^6}-1-3\implies C=\dfrac3{4e^6}$

so that the particular solution is

$y=\dfrac2{1-\frac34e^{2t-6}}-1-t=\boxed{\dfrac8{4-3e^{2t-6}}-1-t}$

3 0

2 years ago