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3 years ago

Find all points having an x-coordinate of 4 whose distance from the point (-2,-1) is 10

1 answer:

8 0

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\
10=\sqrt{(4-(-2))^2+(y-(-1))^2}\\
10=\sqrt{6^2+(y+1)^2}\\
10=\sqrt{36+y^2+2y+1}\\
10=\sqrt{y^2+2y+37}\\
100=y^2+2y+37\\
y^2+2y-63=0\\
y^2+9y-7y-63=0\\
y(y+9)-7(y+9)=0\\
(y-7)(y+9)=0\\
y=7 \vee y=-9\\\\
\boxed{(x,y)=\{(4,-9),(4,7)\}}

$

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12=24-y <br><br><br>help me pleaseee

vitfil [10]

Answer:

y=12

Step-by-step explanation:

12=24-y

step 1

subtract each side by 24

24-24 cancels out

12-24=-12

now we have

-y=-12

we now have to divide by -1

-1/-1=1

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so we can conclude that y=12

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3 years ago

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A college basketball coach has 12 players on his team. eight players are receiving scholarships, and four are not. the coach dec

EleoNora [17]

Solution: The given problem is a binomial distribution.

The probability that a team member receives a scholarship is $\frac{8}{12} = 0.67$

Therefore, the given problem follows binomial with n = 5 and p = 0.6667

Now the probability that 4 of 5 players selected are on scholarship is:

$P(x=4) = \binom{5}{4} \times 0.6667^{4} (1-0.6667)^{5-4}$

$=5 \times 0.1976 \times 0.3333$

$=0.329$

Therefore, the probability that 4 of 5 players selected are on scholarship is 0.329

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3 years ago

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Show that a sequence {sn} coverages to a limit L if and only if the sequence {sn-L} coverages to zero.

Andreyy89

Let ${s_n}_{n\in\Bbb N}$ be a sequence that converges to $L$ . This means for any $\varepsilon>0$ , there is some $N$ such that $|s_n-L|$ for all $n>N$ . From this inequality we see that $|(s_n-L)-0|$ , so it follows that $s_n-L\to0$ .

On the other hand, let ${s_n-L}$ be a sequence that converges to 0. This means $|(s_n-L)-0|$ for all large enough $n$ , and we get the simpler inequality for free, $|s_n-L|$ , so it follows that $s_n\to L$ .