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3 years ago

X^2-10x + y^2 + 6y – 30 = 0

Mathematics

1 answer:

pav-90 [236]3 years ago

4 0

Answer:

(

√

)

(

−

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)

Step-by-step explanation:

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A spacecraft travels at a speed of 5 x 10^4 miles per day. How long will it travel in 1.5 x 10^2 days?

Dimas [21]

Answer:

$7.5 \times 10^6$ miles

Step-by-step explanation:

$(5\times 10^4)\times (1.5 \times 10^2)=\\(5\times 1.5) \times (10^4 \times 10^2)=\\7.5 \times 10^6$

Hope this helps!

6 0

3 years ago

Will give Brainly lollllllll XD......................

marshall27 [118]

Answer:

it looks like it would just be -1

Step-by-step explanation:

there two -x and two positive x which would cancel out. three positive 1 and four negative 1. -4+3= -1

that's the best I could come up with

6 0

2 years ago

Read 2 more answers

Eddie is fishing. He has 1 and one fourth ounces of weight on his line, but his bait isn't getting to the bottom of the lake. Ed

vladimir1956 [14]

1 1/4 + 1/2 + 3/4 =
5/4 + 1/2 + 3/4 =
5/4 + 2/4 + 3/4 =
10/4 = 2 1/2 (or 2.5) oz's <

4 0

3 years ago

Read 2 more answers

24. Find the slope and y-intercept of the line.<br> Y=7/5x-3

Ad libitum [116K]

Answer:

slope = m = 7/5

y-intercept = b = -3

Step-by-step explanation:

y = mx + b

m = slope

b = y-intercept

y = 7/5 x - 3

slope = m = 7/5

y-intercept = b = -3

4 0

1 year ago

Read 2 more answers

Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand

dybincka [34]

Answer: Part a) $P(a)=\frac{1}{\binom{r+w}{r}}$

part b) $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}$

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = $P(E)=\frac{Favourablecases}{TotalCases}$

For part a)

Probability that a red ball is drawn in first attempt = $P(E_{1})=\frac{r}{r+w}$

Probability that a red ball is drawn in second attempt= $P(E_{2})=\frac{r-1}{r+w-1}$

Probability that a red ball is drawn in third attempt = $P(E_{3})=\frac{r-2}{r+w-1}$

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = $P(E_{i})=\frac{r-i}{r+w-i}$

Thus the probability that events $E_{1},E_{2}....E_{i}$ occur in succession is

$P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...$

Thus $P(E)$ = $\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!$

Thus our probability becomes

$P(E)=\frac{1}{\binom{r+w}{r}}$

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at $(r+1)^{th}$ draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals $\binom{r}{r-1}$

Thus the probability becomes $P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}$

Thus required probability of case b becomes $P(E)+ P(E_{i})$

= $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}\\\\$

7 0

3 years ago