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4 years ago

Find the length of EG

Mathematics

2 answers:

sergeinik [125]4 years ago

6 0

The triangles are similar because they have two sides and the angle between them (EFG) respectively equals, and it’s evident the factor between the two triangles is 2. So EG=2CB=2*8=16

Simple!!! Hope it helps!

noname [10]4 years ago

4 0

Answer:

EG = 16

Step-by-step explanation:

Given a segment joining the midpoints of 2 sides , then the segment is half the measure of the third side, thus

EG = 2 × BC = 2 × 8 = 16

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In an arithmetic sequence, the nth term an is given by the formula an=a1+(n−1)d, where a1 is the first term and d is the commo

PilotLPTM [1.2K]

Answer:

105th term of given series is

$a_n=\dfrac{105}{2}$

Step-by-step explanation:

Given series is

$\dfrac{1}{2},\ 1,\ \dfrac{3}{2},\ 2,\ \dfrac{5}{2}.....$

As we can see,

$\textrm{First term},a_1=\dfrac{1}{2}$

Also,

$1-\dfrac{1}{2}=\dfrac{3}{2}-1=2-\dfrac{3}{3}=.....=\dfrac{1}{2}$

hence, we can say given series is in arithmetic progression,

with common difference,

$d=\ \dfrac{1}{2}$

As given in question the nth term in A.P is given by

$a_n=a_1+(n-1)d$

since we have to find the 105th term, so we can write

$a_{105}=\dfrac{1}{2}+(105-1)\dfrac{1}{2}$

$=\dfrac{1}{2}+\dfrac{104}{2}$

$=\dfrac{105}{2}$

Hence, the 105th term of given series of A.P is $\dfrac{105}{2}$ .

7 0

4 years ago

A teacher records the number of students present in her 1st period class each day. This count is a ___________ random variable.

blagie [28]

This count is a discrete random variable (option A).

<h3>What is a discrete random variable?</h3>

A discrete random variable is a variable that contains integers that can only be a limited number of possible values. A discrete random variable is can contain only a finite set of numbers .

An example of discrete random variable is the number of students in the first period class. It is impossible for the number of students in the class to go on indefinitely.

Discrete random variable has the following properties:

It is finite
It is numeric
It is countable
It contains non-negative integers.

A continous random variable is a variable that has an infinite number.

To learn more about discrete data, please check: brainly.com/question/22916429

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2 years ago

Alain throws a stone off a bridge into a river below. The stone's height (in meters above the water), xxx seconds after Alain th

Naya [18.7K]

For this case we have the following function:
$h (x) = - 5x ^ 2 + 10x + 15
$
By the time the stone is thrown, x = 0.
We must evaluate this value of x in the function.
We have then:
$h (0) = - 5 (0) ^ 2 + 10 (0) +15

h (0) = 0 + 0 + 15

h (0) = 15 meters$
Answer:
The height of the stone at the time is thrown is given by:
h (0) = 15 meters

4 0

3 years ago

Read 2 more answers

Please help i dont get it <br>20 POINTSS

Sloan [31]

Answer:

P( not win) = 2/ (n+2)

Step-by-step explanation:

There are n+2 total possibilities

There are n winners

not win = n+2 - n = 2

P( not win) = not win / total

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3 0

4 years ago

Write the trigonometric expression sin(sin−1u−tan−1v) as an algebraic expression in u and v. Assume that the variables u and v r

igomit [66]

Answer:

[u – v√(1 – u²)]/√(1 + v²)

Step-by-step explanation:

Let sin^-1(u) = A, therefore sinA = u.

We know that sin(theta) = opposite/hypothenuse

Therefore, sinA = u/1 and u is the opposite side to angle A while 1 is the hypotenuse. Draw an acute triangle placing u opposite to angle A and 1 as the hypotenuse. By Pythagoras theorem the adjacent would be √(1 – u²).

By doing this, it means cosA = adjacent/hypotenuse = √(1 – u²)/1 = √(1 – u²)

Also, let tan^-1(v) = B, therefore tanB = v.

We know that tan(theta) = opposite/adjacent

Therefore, tanB = v/1 and v is the opposite side to angle B while 1 is the adjacent. Draw an acute triangle placing v opposite to angle B and 1 as the adjacent. By Pythagoras theorem the hypothenuse would be √(1 + v²).

Therefore, sinB = opposite/hypotenuse = v/√(1 + v²) and cosB = adjacent/hypotenuse = 1/√(1 + v²)

Now,

sin[sin^–1(u) – tan^–1(v)] =

sin(A – B) =

sinAcosB – sinBcosA =

u[1/√(1 + v²)] – [v/√(1 + v²)][√(1 – u²)] =

[u/√(1 + v²)] – [v√(1 – u²)/√1 + v²)] =

[u – v√(1 – u²)]/√(1 + v²).

8 0

3 years ago