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asambeis [7]
3 years ago
6

Answer two questions about Systems AAA and BBB: System AAA \text{\quad}start text, end text System BBB \begin{cases}-10x-4y=0\\\

\-2x+7y=8\end{cases} ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ ​ −10x−4y=0 −2x+7y=8 ​ \begin{cases}-10x-4y=8\\\\-2x+7y=0\end{cases} ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ ​ −10x−4y=8 −2x+7y=0 ​ 1) How can we get System BBB from System AAA?
Mathematics
2 answers:
dem82 [27]3 years ago
8 0

Answer:

A

Step-by-step explanation:

and yes

m_a_m_a [10]3 years ago
6 0

Answer:

Its swap out the order of equations and Yes

Step-by-step explanation:

in khan academy

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2) Use a graph to find the length of DE if D(4, -3) and E(-5, -7).​
noname [10]

Answer:

The answer rounded to the nearest tenth is 9.8.

Step-by-step explanation:

The answer itself is really long so I rounded to the tenths place.

3 0
3 years ago
3n-5p+2n=10 solve for n
nikdorinn [45]

Answer:

n=3p

Isolate the variable by dividing each side by factors that don't contain the variable.

5 0
2 years ago
Read 2 more answers
a packe of cards conatins 4 red and 5 black cards. a hand of 5 cards is drawn without replacment. What is the proablity of all f
Vladimir79 [104]

Answer: Our required probability is 0.3387.

Step-by-step explanation:

Since we have given that

Number of red cards = 4

Number of black cards = 5

Number of cards drawn = 5

We need to find the probability of getting exactly three black cards.

Probability of getting a black card = \dfrac{5}{9}

Probability of getting a red card = \dfrac{4}{9}

So, using "Binomial distribution", let X be the number of black cards:

P(X=3)=^5C_3(\dfrac{5}{9})^3(\dfrac{4}{9})^2\\\\P(X=3)=0.3387

Hence, our required probability is 0.3387.

5 0
3 years ago
For the following telescoping series, find a formula for the nth term of the sequence of partial sums
gtnhenbr [62]

I'm guessing the sum is supposed to be

\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}

Split the summand into partial fractions:

\dfrac1{(5k-1)(5k+4)}=\dfrac a{5k-1}+\dfrac b{5k+4}

1=a(5k+4)+b(5k-1)

If k=-\frac45, then

1=b(-4-1)\implies b=-\frac15

If k=\frac15, then

1=a(1+4)\implies a=\frac15

This means

\dfrac{10}{(5k-1)(5k+4)}=\dfrac2{5k-1}-\dfrac2{5k+4}

Consider the nth partial sum of the series:

S_n=2\left(\dfrac14-\dfrac19\right)+2\left(\dfrac19-\dfrac1{14}\right)+2\left(\dfrac1{14}-\dfrac1{19}\right)+\cdots+2\left(\dfrac1{5n-1}-\dfrac1{5n+4}\right)

The sum telescopes so that

S_n=\dfrac2{14}-\dfrac2{5n+4}

and as n\to\infty, the second term vanishes and leaves us with

\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}=\lim_{n\to\infty}S_n=\frac17

7 0
3 years ago
CAN SOMEONE PLEASE HELP!! I DONT UNDERSTAND. i WILL GIVE BRAINLIEST!!!!
HACTEHA [7]

Answer:

u should give me the formula for ur question cuz I kind of understand but I did it years ago

7 0
3 years ago
Read 2 more answers
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