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Valentin [98]
2 years ago
9

Someone smart pls help

Mathematics
1 answer:
OleMash [197]2 years ago
5 0

Answer:

5

Step-by-step explanation:

-2^2=-4

3^2=9

add

-4+9=5

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A card is drawn randomly from a standard deck of cards. you win $10 if the card is a spade or an ace. what is the probability th
klio [65]
The probability of winning is 4/13.

These two events are not mutually exclusive; this means they can happen at the same time.  For two events that are not mutually exclusive,

P(A or B) = P(A) + P(B) - P(A and B)

This gives us 
P(spades or Ace) = P(spades) + P(Ace) - P(spades and Ace)

There are 13 spades out of 52 cards.
There are 4 aces out of 52 cards.
There is 1 card that is a spade and an ace out of 52 cards.

13/52 + 4/52 - 1/52 = 16/52 = 4/13
3 0
3 years ago
Solve the equation for the indicated variable (for q)<br><img src="https://tex.z-dn.net/?f=p%20%3D%20%20%5Cfrac%7Bq%7D%7B4%7D%20
Salsk061 [2.6K]

Answer:

q=4(p-r)

Step-by-step explanation:

p-r=q/4

4(p-r)=q

7 0
3 years ago
Read 2 more answers
Use the laplace transform to solve the given initial-value problem. y' 5y = e4t, y(0) = 2
Basile [38]

The Laplace transform of the given initial-value problem

y' 5y = e^{4t}, y(0) = 2 is  mathematically given as

y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}

<h3>What is the Laplace transform of the given initial-value problem? y' 5y = e4t, y(0) = 2?</h3>

Generally, the equation for the problem is  mathematically given as

&\text { Sol:- } \quad y^{\prime}+s y=e^{4 t}, y(0)=2 \\\\&\text { Taking Laplace transform of (1) } \\\\&\quad L\left[y^{\prime}+5 y\right]=\left[\left[e^{4 t}\right]\right. \\\\&\Rightarrow \quad L\left[y^{\prime}\right]+5 L[y]=\frac{1}{s-4} \\\\&\Rightarrow \quad s y(s)-y(0)+5 y(s)=\frac{1}{s-4} \\\\&\Rightarrow \quad(s+5) y(s)=\frac{1}{s-4}+2 \\\\&\Rightarrow \quad y(s)=\frac{1}{s+5}\left[\frac{1}{s-4}+2\right]=\frac{2 s-7}{(s+5)(s-4)}\end{aligned}

\begin{aligned}&\text { Let } \frac{2 s-7}{(s+5)(s-4)}=\frac{a_{0}}{s-4}+\frac{a_{1}}{s+5} \\&\Rightarrow 2 s-7=a_{0}(s+s)+a_{1}(s-4)\end{aligned}

put $s=-s \Rightarrow a_{1}=\frac{17}{9}$

\begin{aligned}\text { put } s &=4 \Rightarrow a_{0}=\frac{1}{9} \\\Rightarrow \quad y(s) &=\frac{1}{9(s-4)}+\frac{17}{9(s+s)}\end{aligned}

In conclusion, Taking inverse Laplace tranoform

L^{-1}[y(s)]=\frac{1}{9} L^{-1}\left[\frac{1}{s-4}\right]+\frac{17}{9} L^{-1}\left[\frac{1}{s+5}\right]$ \\\\

y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}

Read more about Laplace tranoform

brainly.com/question/14487937

#SPJ4

6 0
1 year ago
Three times a number t minus twelve equals forty
tekilochka [14]
3 · t - 12 = 40

3t - 12 = 40    |add 12 to both sides

3t = 52    |divide both sides by 3

t = 52/3

t = 17 1/3
5 0
3 years ago
About 250 million drivers are registered in the United States. While on a trip with her family, a student counted the number of
Lilit [14]
She could estimate that there would be about 148 million female drivers.

From her data, she found there were 118 females drivers out of the total 200 drivers that she counted.

That is 118 out of 200 or 0.59 or 59%.

Now, just take 59% of 250.

0.59 x 250 = 147.5 which rounds to 148.
4 0
3 years ago
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