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Find all solutions for a triangle with a = 14, b= 12, and B = 90°.

ziro4ka [17]

Answer:

No solution.

Step-by-step explanation:

Given

$a = 14; b= 12; B = 90^\circ$

Required

Find all possible solutions

If $B = 90^\circ$ , then the triangle is right-angled and the hypotenuse is at b

Given that $a = 14; b= 12;$

This implies that a > b

In a right-angled triangle, the hypotenuse (side b) is the longest.

Since this is not true for the given sides, then the triangle has no solution.

5 0

3 years ago

Wyatt makes and sells rope doormats. He uses 200 feet of rope for each doormat.

skad [1K]

Correct answer is $200(25)(\frac{1}{3})$ or , 200×25÷3 . Given answer is not in any of option! kindly match this answer with your options available!

<u>Step-by-step explanation:</u>

Here we have , Wyatt makes and sells rope doormats. He uses 200 feet of rope for each doormat. He wants to make 25 doormats. We need to find Which expression shows a correct way to find how many yards of rope he needs. Let's find out:

Wyatt makes and sells rope doormats. He uses 200 feet of rope for each doormat , i.e. for 1 doormat he uses 200 feet of rope

⇒ $1 Doormat=200 ft$

So , for 25 Doormat he uses :

⇒ $25(1 Doormat)=25(200) ft$

⇒ $25 Doormat=25(200) ft$

But , $1 yard = 3 ft$ ⇒ $1 feet = \frac{1}{3} yard$ .Hence ,

⇒ $25 Doormat=25(200) (\frac{1}{3})yards$

Hence Correct answer is $200(25)(\frac{1}{3})$ or , 200×25÷3 . Given answer is not in any of option! kindly match this answer with your options available!

7 0

4 years ago

Example 2: Graph 2x-5y210

salantis [7]

Answer:

I am not sure this is graphable

Step-by-step explanation:

There needs to be an = sign

3 0

3 years ago

He gets 1/8 of his sales as a connission. How much must he sell in order to earn $500?

Licemer1 [7]

Answer:

4000

Step-by-step explanation:

500x8=4000

8 0

3 years ago

The height of a stuntperson jumping off a building that is 20 m high is modeled by the equation h = 20 – 5t2, where t is the tim

Leno4ka [110]

<h2>Time Interval : 1 sec to 1.414 sec</h2>

Step-by-step explanation:

The height of the stuntperson jumping off a $20\text{ }m$ is modelled as $h\text{ }=\text{ }20\text{ }-\text{ }5t^{2}$ . Here $t$ is the time in seconds.

We have to record the person when he is between the heights $15\text{ }m\text{ and }10\text{ }m$ from the ground.

When $h\text{ }=\text{ }15$ ,

$15\text{ }=20\text{ }-\text{ }5t^{2}\\5t^{2}\text{ }=\text{ }5\\t^{2}\text{ }=\text{ }1\\t\text{ }=\text{ }1\text{ }sec$

When $h\text{ }=\text{ }10$ ,

$10\text{ }=20\text{ }-\text{ }5t^{2}\\5t^{2}\text{ }=\text{ }10\\t^{2}\text{ }=\text{ }2\\t\text{ }=\text{ }\sqrt{2}\text{ }=\text{ }1.414\text{ }sec$

The stuntperson is at height $15\text{ }m$ at $t\text{ }=\text{ }1\text{ }sec$ , and is at height $10\text{ }m$ at $t\text{ }=\text{ }1.414\text{ }sec$ .

∴ The camera should film the person for the time interval $\textbf{1}\text{ }sec\text{ to }\textbf{1.414}\text{ }sec$ .

5 0

3 years ago

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