Ask question

Ask question

All categories

4 years ago

10

How do i find LM? On number 2 .

1 answer:

Masja [62]4 years ago

4 0

Given: KL is a tangent to the circle.
LM is another tangent to the circle.

We can use the tangent meeting at an external point theorem.
<em>Theorem</em>: <em>The tangent segments to a circle from an external point are equal.
</em>
Thus, we can say KL = LM, as they lie on a common circle, and are tangents to such circle.

4x - 2 = 3x + 3
4x - 3x = 3 + 2
x = 5

Since LM is 3x + 3, we can substitute the value of x to LM to render:
3(5) + 3 = 18

Thus, LM = KL = 18 units.

You might be interested in

5r-2=6r+5 what does r equal

Anuta_ua [19.1K]

Answer:

r=-7

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

The ratio between the number of students (s) on a field trip and the number of teachers (t) on the field trip is 21/4. There are

RoseWind [281]

Answer:

Step-by-step explanation:

If there needs to be 4 teachers per 21 students, you would divide the total amount of students by 21 and multiply that answer by 4. SO...:

84 ÷ 21 = 4 groups of 21 students.

4 groups × 4 teachers = 16

*** Answer***

16 TEACHERS

5 0

3 years ago

Quadrilateral ABCD shown below, is translated 3 units to the left to create quadrilateral A'BCD

Anton [14]

I believe b but I could be so wrong I’m sorry if I am

3 0

3 years ago

A fair coin is tossed three times and the events A, B, and C are defined as follows: A:{ At least one head is observed } B:{ At

kicyunya [14]

Answer:

(a) 1/2

(b) 1/2

(c) 1/8

Step-by-step explanation:

Since, when a fair coin is tossed three times,

The the total number of possible outcomes

n(S) = 2 × 2 × 2

= 8 { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT },

Here, B : { At least two heads are observed } ,

⇒ B = {HHH, HHT, HTH, THH},

⇒ n(B) = 4,

Since,

$\text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}$

(a) So, the probability of B,

$P(B) =\frac{n(B)}{n(S)}=\frac{4}{8}=\frac{1}{2}$

(b) A : { At least one head is observed },

⇒ A = {HHH, HHT, HTH, THH, HTT, THT, TTH},

∵ A ∩ B = {HHH, HHT, HTH, THH},

n(A∩ B) = 4,

$\implies P(A\cap B) = \frac{n(A\cap B)}{n(S)} = \frac{4}{8}=\frac{1}{2}$

(c) C: { The number of heads observed is odd },

⇒ C = { HHH, HTT, THT, TTH},

∵ A ∩ B ∩ C = {HHH},

⇒ n(A ∩ B ∩ C) = 1,

$\implies P(A\cap B\cap C)=\frac{1}{8}$

7 0

3 years ago

PLEASE HELP ASAP!! WILL GIVE BRAINLIEST!!

zhenek [66]

Given is :

The area of the rectangular patio is = 130 square feet

Let the width of the patio be = x feet

As length is 3 feet more than the width, so the length is = x+3 feet

Now area of rectangle is = length * width

$x(x+3)=130$

$x^{2}+3x=130$

$x^{2}+3x-130=0$

$x^{2}+13x-10x-130=0$

$x(x+13)-10(x+13)=0$

$(x-10)(x+13)=0$

This gives either x-10=0 or x+13=0

x-10=0 ; x=10

x+13=0 ; x=-13 (neglect this negative value)

So, the width of the patio is = 10 feet

Length = 10+3 = 13 feet

7 0

3 years ago