Ask question

Ask question

All categories

3 years ago

11

Can the sides of a triangle have lengths 3, 3, and 10?

2 answers:

Bingel [31]3 years ago

8 0

Answer:

No

Step-by-step explanation:

According to the property of the triangle ,

sum of any two sides should we greater then third side of the triangle.

Here, Measurement of three sides are given as 3,3,5 .

So, sum of the measurement of first two sides is 6.

And third side equals 10.

Clearly 6 is less than 10. So . it violates the property sum of any two sides should we greater then third side of the triangle.

Thus , Sides of a triangle can't be 3,3,10.

Igoryamba3 years ago

3 0

Answer:

Yes

Step-by-step explanation:

As long as there are three sides and it forms a triangle, the answer is yes.

You might be interested in

Weary of the low turnout in student elections, a college administration decides to choose an SRS of three students to form an ad

-Dominant- [34]

Answer:

$P(ABC) = 0.110592$

$P(ABC^c) = 0.119808$

$P(AB^cC) = 0.119808$

$P(A^cBC) = 0.119808$

$P(AB^cC^c) = 0.129792$

$P(A^cBC^c) = 0.129792$

$P(A^cB^cC) = 0.129792$

$P(A^cB^cC^c) = 0.140608$

Step-by-step explanation:

Given

$P(A) = P(B) = P(C) = 48\%$

Convert the probability to decimal

$P(A) = P(B) = P(C) = 0.48$

Solving (a): P(ABC)

This is calculated as:

$P(ABC) = P(A) * P(B) * P(C)$

This gives:

$P(ABC) = 0.48*0.48*0.48$

$P(ABC) = 0.110592$

Solving (b): $P(ABC^c)$

This is calculated as:

$P(ABC^c) = P(A) * P(B) * P(C^c)$

In probability:

$P(C^c) = 1 - P(C)$

So, we have:

$P(ABC^c) = P(A) * P(B) * (1 - P(C))$

$P(ABC^c) = 0.48 * 0.48 * (1 - 0.48)$

$P(ABC^c) = 0.48 * 0.48 * 0.52$

$P(ABC^c) = 0.119808$

Solving (c): $P(AB^cC)$

This is calculated as:

$P(AB^cC) = P(A) * P(B^c) * P(C)$

$P(AB^cC) = P(A) * [1 - P(B)] * P(C)$

$P(AB^cC) = 0.48 * (1 - 0.48)* 0.48$

$P(AB^cC) = 0.48 * 0.52* 0.48$

$P(AB^cC) = 0.119808$

Solving (d): $P(A^cBC)$

This is calculated as:

$P(A^cBC) = P(A^c) * P(B) * P(C)$

$P(A^cBC) = [1-P(A)] *P(B) * P(C)$

$P(A^cBC) = (1 - 0.48)* 0.48 * 0.48$

$P(A^cBC) = 0.52* 0.48 * 0.48$

$P(A^cBC) = 0.119808$

Solving (e): $P(AB^cC^c)$

This is calculated as:

$P(AB^cC^c) = P(A) * P(B^c) * P(C^c)$

$P(AB^cC^c) = P(A) * [1-P(B)] * [1-P(C)]$

$P(AB^cC^c) = 0.48 * [1-0.48] * [1-0.48]$

$P(AB^cC^c) = 0.48 * 0.52*0.52$

$P(AB^cC^c) = 0.129792$

Solving (f): $P(A^cBC^c)$

This is calculated as:

$P(A^cBC^c) = P(A^c) * P(B) * P(C^c)$

$P(A^cBC^c) = [1-P(A)] * P(B) * [1-P(C)]$

$P(A^cBC^c) = [1-0.48] * 0.48 * [1-0.48]$

$P(A^cBC^c) = 0.52 * 0.48 * 0.52$

$P(A^cBC^c) = 0.129792$

Solving (g): $P(A^cB^cC)$

This is calculated as:

$P(A^cB^cC) = P(A^c) * P(B^c) * P(C)$

$P(A^cB^cC) = [1-P(A)] * [1-P(B)] * P(C)$

$P(A^cB^cC) = [1-0.48] * [1-0.48] * 0.48$

$P(A^cB^cC) = 0.52 * 0.52 * 0.48$

$P(A^cB^cC) = 0.129792$

Solving (h): $P(A^cB^cC^c)$

This is calculated as:

$P(A^cB^cC^c) = P(A^c) * P(B^c) * P(C^c)$

$P(A^cB^cC^c) = [1-P(A)] * [1-P(B)] * [1-P(C)]$

$P(A^cB^cC^c) = [1-0.48] * [1-0.48] * [1-0.48]$

$P(A^cB^cC^c) = 0.52*0.52*0.52$

$P(A^cB^cC^c) = 0.140608$

5 0

2 years ago

jesse walks 3 miles per hour to school.the school is 1.5 miles away.How long would it take jesse to walk to school

Pavlova-9 [17]

1 hour 30 minutes!
Please mark this the brainlest answer

5 0

2 years ago

Find the equation of the line through point (2,−4) and parallel to −6x+2y=4. Use a forward slash (i.e. "/") for fractions (e.g.

chubhunter [2.5K]

Answer:

y = 3x - 10

Step-by-step explanation:

Step 1: Rewrite 1st equation

2y = 6x + 4

y = 3x + 4

Step 2: Find 2nd equation

y = 3x + b

-4 = 3(2) + b

-4 = 6 + b

b = -10

Step 3: Rewrite 2nd equation

y = 3x - 10

And we have your parallel equation!

5 0

3 years ago

How could you model 5 students sharing 4 bags of popcorn equally? How much will each student get?

qwelly [4]

One and a fourth bags of popcorn

4 0

3 years ago

A line passes through the point (2, -9) and has a slope of -9. Write an equation in point-slope form for this line.

Natali [406]

Answer:

$y+9 = - 9x +18$

Step-by-step explanation:

Line is passing through the point $(2,\: - 9) =(x_1,\:y_1)$ and has slope (m) = - 9

General equation of line in point slope form is given as:

$y-y_1 = m(x-x_1)$

Therefore

$y-(-9) = - 9(x-2)$

$y+9 = - 9x +18$

This is the required equation of line in point slope form.

4 0

3 years ago