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zhenek [66]
3 years ago
14

2357111317 what is the next number

Mathematics
1 answer:
ArbitrLikvidat [17]3 years ago
7 0
The next number will be 19 since you keep adding 2 each time
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A different class has 285 students, and 51.2% of them are men. how many men are in the class?​
Zanzabum

Answer:

146

Step-by-step explanation:

Get a percent calculator, Then, you put 51.2% of 285 and you actually get 145.96 but the rounded of what you need is 146

6 0
3 years ago
Simplify: X + x = <br>5y + y =​
Oduvanchick [21]

Answer:

x + x = 2x \\ 5y + y = 6y \\ thank \: ou

8 0
2 years ago
Read 2 more answers
Find a particular solution to the nonhomogeneous differential equation y??+4y?+5y=?10x+e^(?x).
Firdavs [7]

Answer:

A) Particular solution:

2x+\frac{1}{2}e^{-x}-\frac{8}{5}

B) Homogeneous solution:

y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))

C) The most general solution is

y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}

Step-by-step explanation:

Given non homogeneous ODE is

y''+4y'+5y=10x+e^{-x}---(1)

To find homogeneous solution:

D^{2}+4D+5=0\\D^{2}+4D+4-4+5=0\\\\(D+2)^{2}=-1\\D+2=\pm iD=-2 \pm i\\y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))---(2)

To find particular solution:

y_{p}=Ax+B+Ce^{-x}\\\\y'_{p}=A-Ce^{-x}\\y''_{p}=Ce^{-x}\\

Substituting y_{p},y'_{p},y''_{p} in (1)

y''_{p}+4y'_{p}+5y_{p}=10x+e^{-x}\\Ce^{-x}+4(A-Ce^{-x})+5(Ax+B+Ce^{-x})=10x+e^{-x}\\

Equating the coefficients

5Ax+2Ce^{-x}+4A+5B=10x+e^{-x}\\5A=10\\A=2\\4A+5B=0\\B=-\frac{4A}{5}B=-\frac{8}{5}2C=1\\C=\frac{1}{2}\\So,\\y_{p}=2x+\frac{1}{2}e^{-x}-\frac{8}{5}---(3)\\

The general solution is

y=y_{h}+y_{p}

from (2) ad (3)

y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}

6 0
2 years ago
karen drove 3/4 of a mile in 1/60 of an hour .At this rate,how many miles would she drive in one hour
amid [387]

Answer:

45 miles

Step-by-step explanation:

1/60 of one hour (60 minutes) equates to 1 minute

Karen is driving 3/4 of a mile in 1 minute to multiply 3/4 and 60 to get 45 miles

6 0
3 years ago
A projectile is launched into the air. The function h(t) = –16t2 + 32t + 128 gives the height, h, in feet, of the projectile t s
Zinaida [17]

Answer:

t = 4 seconds

Step-by-step explanation:

The height of the projectile after it is launched is given by the function :

h(t)=-16t^2+32t+128

t is time in seconds

We need to find after how many seconds will the projectile land back on the ground. When it land, h(t)=0

So,

-16t^2+32t+128=0

The above is a quadratic equation. It can be solved by the formula as follows :

t=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}

Here, a = -16, b = 32 and c = 128

t=\dfrac{-32\pm \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}\\\\t=\dfrac{-32+ \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}, \dfrac{-32\- \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}\\\\t=-2\ s\ \text{and}\ 4\ s

Neglecting negative value, the projectile will land after 4 seconds.

4 0
3 years ago
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