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Vikentia [17]
3 years ago
12

A triangle has two sides of lengths 4 and 5. What value could the length of the third side be?

Mathematics
2 answers:
Olin [163]3 years ago
6 0

Answer:

2, 3, 4, 5, 6, 7 or 8.

Step-by-step explanation:

We know that the sum of two sides on a triangle should ALWAYS be greater than the third side. Then we have:

5-4 < x < 5 + 4

1 < x < 9

Therefore, the lenght of the third side could be any number between 1 and 9. If the lenght of the third side is an integrer, then the lenght could be:

2, 3, 4, 5, 6, 7 or 8.

ANTONII [103]3 years ago
4 0

Answer:

The length of the third side could be all real numbers greater than 1 unit and less than 9 units

Step-by-step explanation:

we know that

The <u>Triangle Inequality Theorem,</u>   states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side

so

Applying the triangle inequality theorem

Let

x ----> the length of the third side

1) 4+5 > x

9 > x

Rewrite

x < 9 units

2) 4+x > 5

x > 5-4

x > 1 units

therefore

The solution for the third side is the interval -----> (1,9)

All real numbers greater than 1 unit and less than 9 units

therefore

The length of the third side could be all real numbers greater than 1 unit and less than 9 units

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A roulette wheel has 38 3838 slots, of which 18 1818 are red, 18 1818 are black, and 2 22 are green. In each round of the game,
Elina [12.6K]

The value of P(R = 3) is 0.2854

<h3>Probability</h3>

Probabilities are used to determine the chances of events.

The given parameters are:

  • n = 38 --- the number of slots
  • red = 18. --- the number of red slot
  • black = 18. --- the number of black slot
  • green = 2. --- the number of green slot

<h3>Individual probabilities</h3>

The probability that the ball lands on a red slot is calculated as:

p = \frac{18}{38}

Simplify

p= \frac{9}{19}

The probability that the ball does not land on a red slot is calculated as:

q = \frac{18+2}{38}

q = \frac{20}{38}

Simplify

q = \frac{10}{19}

<h3>Binomial probability</h3>

The value of P(R = 3) is then calculated using the following binomial probability formula

P(R = r) = ^nC_rp^rq^{n-r}

Where:

n = 7

r = 3

So, we have:

P(R = 3) = ^7C_3 \times (9/19)^3 \times (10/19)^{7-3}

P(R = 3) = 35 \times (9/19)^3 \times (10/19)^4

P(R = 3) = 0.2854

Hence, the value of P(R = 3) is 0.2854

Read more about probabilities at:

brainly.com/question/15246027

8 0
2 years ago
Read 2 more answers
Ill give brainlist if you can answer this for me
NeTakaya

Answer:

The third one

Step-by-step explanation:

consider the following image

\frac{10}{\sqrt{2} } =5\sqrt{2}

4 0
3 years ago
Angela enjoys swimming and often swims at a steady pace to burn calories. At this pace, Angela can swim 1,700 meters in 40 minut
KATRIN_1 [288]
Well, if you are talking how much she can swim per minute, you have to divide 1700 by 40 to get the unit per minute, 1700 being the meters she can swim, and 40 are the minutes she takes to swim. So, 1700 divided by 40 would be 42.5 meters per minute. So Angela's unit is 42.5 meters per minute. YOU'RE WELCOME :D
3 0
3 years ago
Please show me the work!
Papessa [141]

Answer:

9/2π cm/s

Step-by-step explanation:

Volume of the spherical balloon V= 4/3πr³

Rate of change of volume dV/dt = dV/dr * dr/dt

dV/dr = 3(4/3πr²)

dV/dr = 4πr²

dV/dr = 4π(2)²

dV/dr = 16π

If V = 72cm²/s

72 = 16π*dr/dt

dr/dt = 72/16π

dr/dt = 9/2π cm/s

Hence the radius is changing at  9/2π cm/s

6 0
3 years ago
Coefficient of b in expansion of (3+b)^4
Margarita [4]
Are you sure you want ONLY the coefficient of b?  If you expand this, you will have b in 3 of 4 terms.

According to Pascal's Triangle, the coefficients of (a+b)^4 are as follows:

                             1
                        1    2    1
                   1      3    3     1
              1       4     6     4     1

So (a+b)^4 would be 1a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4

Here, you want (3 + b)^4.  Here's what that looks like:

3^4 + 4[3^3*b] + 6[3^2*b^2] + 4[3*b^3] + 1[b^4]

Which coeff did you want?
4 0
3 years ago
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