Ask question

Ask question

All categories

3 years ago

12

MK and ML are tangent to circle J. If MK = 6x – 15 and ML = 2x + 9, find x. * am i right?? * --> provide details if i'm wrong

.

1 answer:

Veseljchak [2.6K]3 years ago

3 0

Answer:

Step-by-step explanation:

$y^{2} = r^{2} + (4,2)^{2} \\\\ y^{2} - r^{2} = (4,2)^{2} y^{2} = r^{2} + z^{2} \\\\ y^{2} - r^{2} = z^{2}\\so: z^{2} = (4,2)^{2} ----> z = 4,2$

You might be interested in

Convert.<br> {} {}<br> minutes ==equals 888 hours 373737 minutes

soldi70 [24.7K]

9514 1404 393

Answer:

517 minutes

Step-by-step explanation:

There are 60 minutes in an hour, so 8×60 = 480 minutes in 8 hours.

In 8 hours 37 minutes, there are ...

480 min + 37 min = 517 minutes

5 0

3 years ago

Write the formula for a function d(x) that describes the distance between the point P and a point (x,y) on the line. You final a

vlabodo [156]

Answer:

d(x) = √[(x - 2)² + (3x - 1)²]

Step-by-step explanation:

The distance between two points with coordinates (x₁, y₁) and (x₂, y₂) is given as

d = √[(x₂ - x₁)² + (y₂ - y₁)²]

So, the distance between point (2,0) and a point (x,y)

d = √[(x - 2)² + (y - 0)²]

d = √[(x - 2)² + (y)²]

But the point (x,y) is on the line y = 3x - 1

We can substitute for y in the distance between points equation.

d(x) = √[(x - 2)² + (3x - 1)²]

QED!

6 0

3 years ago

One hundred items are simultaneously put on a life test. Suppose the lifetimes

romanna [79]

Answer:

a) $\mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}$

b) $\mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}$

Step-by-step explanation:

Given:

The lifetimes of the individual items are independent exponential random variables.

Mean = 200 hours.

Assume, Ti be the time between ( $i-1$ )st and the $ith$ failures. Then, the $T_{i}$ are independent with $\mathrm{T}_{\mathrm{i}}$ being exponential with rate $\frac{(101-i)}{200} .$

Therefore,

a) $E[T]=\sum_{i=1}^{5} E\left[\tau_{i}\right]$

$=\sum_{i=1}^{5} \frac{200}{101-i}$

$\therefore \mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}$

$b)$

The variance is given by, $\mathrm{Var}[\mathrm{T}]=\sum_{i=1}^{5} \mathrm{Var}[T]$

$\therefore \mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}$

7 0

3 years ago

Hi i would like sum help :)) please help for a brainlist !!!!!!!!!!!!!

Sergio [31]

A. all real numbers for both

Explanation:
Domain is the x-values. Since the graph goes forever horizontally, any x-value is on the graph.
Range is the same but with y-values. If the graph went on further than what the picture shows, then it also would show that every y value is on the graph
hope this helps

5 0

3 years ago

What are the solutions of the quadratic equation? 4x^2+34x+60=0

insens350 [35]

Answer:

Option 3) -6,-5/2

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$4x^{2} +34x+60=0$

so

$a=4\\b=34\\c=60$

substitute in the formula

$x=\frac{-34(+/-)\sqrt{34^{2}-4(4)(60)}} {2(4)}$

$x=\frac{-34(+/-)\sqrt{196}} {8}$

$x=\frac{-34(+/-)14} {8}$

$x_1=\frac{-34(+)14} {8}=-\frac{20}{8}=-\frac{5}{2}$

$x_2=\frac{-34(-)14} {8}=-6$

6 0

2 years ago