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3 years ago

What is the distance between (6, 1) and (1, -9)?

Mathematics

1 answer:

AVprozaik [17]3 years ago

4 0

Answer: √15 units

Step-by-step explanation:

Let (6,1) be (x^1,y^1) and (1,-9) be (x^2,y^2) .

As we know ,

Distance(D) = √(x^1-x^2) +(y^1-y^2)

Now,

D= √(x^1-x^2) +(y^1-y^2)

= √(6-1) +(1+9)

= √5+10

= √15 units

: Therefore the distance between (6,1) and (1,-9) is √15 units.

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the axis of symmetry of a quadratic equation is x = –3. if one of the zeroes of the equation is 4, what is the other zero?

FinnZ [79.3K]

ANSWER

The other zero is

$x = - 10$

EXPLANATION

The axis of symmetry serves as the midpoint of the two zeroes.

We were given that the axis of symmetry of the quadratic equation is
$x = - 3$

We were also given that, one of the zeroes of the quadratic equation is
$x = 4$

Let the other zero of the quadratic equation be
$x = p$

,then, we apply the midpoint formula to find the value of
$p$

Since it is the x-values of the intercepts that gives the solution, we use only the x-value part of the midpoint formula which is given by,

$\frac{x_1 + x_2}{2} = axis \: of \: symmetry$

We substitute to obtain,

$\frac{p + 4}{2} = - 3$

We multiply through by 2 to get,

$p + 4 = - 6$

This implies that,

$p = - 6 - 4$

$p = - 10$

3 0

3 years ago

Read 2 more answers

Alice searches for her term paper in her filing cabinet, which has several drawers. She knows thatshe left her term paper in dra

katen-ka-za [31]

You made a mistake with the probability $p_{j}$ , which should be $p_{i}$ in the last expression, so to be clear I will state the expression again.

So we want to solve the following:

Conditioned on this event, show that the probability that her paper is in drawer $j$ , is given by:

(1) $\frac{p_{j} }{1-d_{i}p_{i} } , if j \neq i,$ and

(2) $\frac{p_{i} (1-d_{i} )}{1-d_{i}p_{i} } , if j = i.$

so we can say:

$A$ is the event that you search drawer $i$ and find nothing,

$B$ is the event that you search drawer $i$ and find the paper,

$C_{k}$ is the event that the paper is in drawer $k, k = 1, ..., n.$

this gives us:

$P(B) = P(B \cap C_{i} ) = P(C_{i})P(B | C_{i} ) = d_{i} p_{i}$

$P(A) = 1 - P(B) = 1 - d_{i} p_{i}$

Solution to Part (1):

if $j \neq i$ , then $P(A \cap C_{j} ) = P(C_{j} )$ ,

this means that

$P(C_{j} |A) = \frac{P(A \cap C_{j})}{P(A)} = \frac{P(C_{j} )}{P(A)} = \frac{p_{j} }{1-d_{i}p_{i} }$

as needed so part one is solved.

Solution to Part(2):

so we have now that if $j$ = $i$ , we get that:

$P(C_{j}|A ) = \frac{P(A \cap C_{j})}{P(A)}$

remember that:

$P(A|C_{j} ) = \frac{P(A \cap C_{j})}{P(C_{j})}$

this implies that:

$P(A \cap C_{j}) = P(C_{j}) \cdot P(A|C_{j}) = p_{i} (1-d_{i} )$

so we just need to combine the above relations to get:

$P(C_{j}|A) = \frac{p_{i} (1-d_{i} )}{1-d_{i}p_{i} }$

as needed so part two is solved.

8 0

4 years ago

Plz solve for the area, questions 10,11,and 12 plz

marishachu [46]

Answer:

10-52.85ft squared 11-144.84centimeters squared 12-304meters squared

Step-by-step explanation:

A-bh/2 10-7*15.1=105.7/2=52.85ft squared

A-bh 11-10.2*14.2=144.84cenitmeters squared

A-bh 20*15.2=304meters squared

4 0

4 years ago

Simplify 7/10 - 1/10

RSB [31]

Answer:3/5

Step-by-step explanation:

6 0

4 years ago

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Two angles are supplementary. the measure of one angle is 4 more than 2 times the measure of teh other andgle. Write an equation

ch4aika [34]

Answer:

$y=58.67^{\circ}$

$x=121.33^{\circ}$

Step-by-step explanation:

We are given that two angles are supplementary .

We have to write an equation that can be used to find the measures.

Let x and y are supplementary

According to question

$x+y=180^{\circ}$ ( by definition of supplementary angles)

$x=2y+4$

Substitute the value then we get

$2y+4+y=180$

$3y=180-4$

$3y=176$

$y=\frac{176}{3}$

$y=58.67^{\circ}$

Substitute the value then, we get

$x+58.67=180$

$x=180-58.67$

$x=121.33^{\circ}$

3 0

4 years ago

What is the distance between (6, 1) and (1, -9)?​

What is the distance between (6, 1) and (1, -9)?