All categories

3 years ago

The measures of two sides of a triangle are 4in and 13in. Find all possible values that the third side of this triangle can be?

Mathematics

1 answer:

ElenaW [278]3 years ago

5 0

Answer:

9 < x < 17 is the possible length of the third side of a triangle.

Step-by-step explanation:

The Triangle Inequality theorem defines that if we are given two sides of a triangle, the sum of any two given sides of a triangle must be greater than the measure of the 3rd side.

Given the two sides of the triangle

4 in

13 in

Let 'x' be the length of 3rd size.

According to the Triangle Inequality theorem,

The difference of two sides < x < The sum of two sides

13 - 4 < x < 13+4

9 < x < 17

Therefore, 9 < x < 17 is the possible length of the third side of a triangle.

You might be interested in

How many lists of positive consecutive integers sum to 100?

hjlf

I found two using a little C# program:

9,10,11,12,13,14,15,16

and

18,19,20,21,22

(if you're interested in the program drop me a msg)

7 0

3 years ago

What are the like terms in the expression 7y + 4 +4x + 7

lorasvet [3.4K]

Answer:

4, 7

Step-by-step explanation:

7y + 4 +4x + 7

The only like terms are the constants 4 and 7

7y is like yogurt and 4x is like xrays, we don't mix them

7 0

3 years ago

Read 2 more answers

Prove :

Sauron [17]

Answer:

See Below.

Step-by-step explanation:

We want to verify the equation:

$\displaystyle \frac{1}{\sec\alpha+1}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

We can convert sec(α) to 1 / cos(α):

$\displaystyle \frac{1}{1/\cos\alpha+1}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

Multiply both layers of the first fraction by cos(α):

$\displaystyle \frac{\cos\alpha}{1+\cos\alpha}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

Create a common denominator. We can multiply the first fraction by (1 - cos(α)):

$\displaystyle \frac{\cos\alpha(1-\cos\alpha)}{(1+\cos\alpha)(1-\cos\alpha)}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

Simplify:

$\displaystyle \frac{\cos\alpha(1-\cos\alpha)}{1-\cos^2\alpha}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

From the Pythagorean Identity, we know that cos²(α) + sin²(α) = 1 or equivalently, 1 - cos²(α) = sin²(α). Substitute:

$\displaystyle \frac{\cos\alpha(1-\cos\alpha)}{\sin^2\alpha}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

Subtract:

$\displaystyle \frac{\cos\alpha(1-\cos\alpha)-\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Distribute:

$\displaystyle \frac{\cos\alpha-\cos^2\alpha-\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Rewrite:

$\displaystyle \frac{(\cos\alpha)-(\cos^2\alpha+\cos\alpha)}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Split:

$\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{\cos^2\alpha+\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Factor the second fraction, and substitute sin²(α) for 1 - cos²(α):

$\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{\cos\alpha(\cos\alpha+1)}{1-\cos^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Factor:

$\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{\cos\alpha(\cos\alpha+1)}{(1-\cos\alpha)(1+\cos\alpha)}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Cancel:

$\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{\cos\alpha}{(1-\cos\alpha)}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Divide the second fraction by cos(α):

$\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}=\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Hence proven.

7 0

2 years ago

In the figure, is an exterior angle to triangle .

ludmilkaskok [199]

Answer:

I don't have c but I do have a and b

Step-by-step explanation:

Part A)

Angle 3 is 103° because 77° + ∠3= 180° and ∠3 = 180° - 77° = 103°

Part B)

Angle 4 is 77° because ∠3 + ∠4 = 180° and ∠4 = 180° - 103° = 77°

These are the answers for k12

3 0

2 years ago

Solve for n. c/d = d/c - 1/n

Sedaia [141]

Answer:

Hi! The other answer is not correct to your question

The answer to your question is n= $\frac{dc}{(c+d)(c-d)}$

Step-by-step explanation:

Solve the rational equation by combining expressions and isolating the variable n

※※※※※※※※※※※※※※※

⁅Brainliest is greatly appreciated!⁆

- Brooklynn Deka

Hope this helps!!

6 0

3 years ago