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3 years ago

Can the side lengths 12,15, and 13 form a triangle?

Mathematics

2 answers:

Fantom [35]3 years ago

8 0

Answer:

Yes

Step-by-step explanation:

It is a Pythagorean Triple. A Pythagorean Triple is the sides of a triangle that fit perfectly into the Pythagorean Theorem, which is:

a²+b²=c²

12, 13, and 15 are numbers that you should know off the top of your head so if you see a triangle with 2 of those numbers, you instantly know the 3rd number.

~Stay golden~ :)

Ronch [10]3 years ago

3 0

Answer: Yes

Step-by-step explanation: To determine if these side lengths can form a triangle, I attached a rule in the image provided which is very helpful to look at especially when you're new at this.

If a triangle has sides with lengths of 12, 15, and 13, notice that 12 + 15 or 27 is greater than 13.

So the sum of the lengths of two sides of the triangle is greater than the length of the third side.

This means that the triangle with sides of lengths of 12, 15, and 13, is possible.

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Enter the correct answer in the box. What is the inverse of function f? f(x)=-7x-4

Naddika [18.5K]

Answer:

The inverse is -1/7(x+4)

Step-by-step explanation:

We have the function

y = -7x-4

To find the inverse, exchange x and y

x = -7y -4

Solve for y

Add 4 to each side

x+4 = -7y -4+4

x+4 = -7y

Divide each side by -7

-1/7(x+4) = -7y/-7

-1/7(x+4) = y

The inverse is -1/7(x+4)

8 0

2 years ago

If 6x^4+8x^3-5x^2+ax+b is completely divisible by 2x^2-5 then find the value of a and b.

mixas84 [53]

Answer:

The value of $a = -20$

The value of $b=-25$

Step-by-step explanation:

Let

$p(x)=6x^4+8x^3-5x^2+ax+b$

and

$q(x)=2x^2-5$

As $q(x)$ is exactly divisible, so, the remainder is 0.

∴ $p(x)=q(x).g(x)+r(x)$

$6x^4 + 8x^3 - 5x^2 + ax + b = (\sqrt{2} x - \sqrt{5})(\sqrt{2} x + \sqrt{5}) g(x)$

As the R.H.S becomes 0 when $x = \sqrt{\frac{5}{2} }$ and $x = -\sqrt{\frac{5}{2} }$ .

And

by substituting $x = \sqrt{\frac{5}{2} }$

$6(\sqrt{\frac{5}{2} } )^4 + 8(\sqrt{\frac{5}{2} } )^3 - 5(\sqrt{\frac{5}{2} } )^2 + a(\sqrt{\frac{5}{2} } ) + b = 0$

$25 + 20( \frac{\sqrt{5}}{\sqrt{2}} ) + a( \frac{\sqrt{5}}{\sqrt{2}}) +b =0...[A]$

by substituting $x = -\sqrt{\frac{5}{2} }$

$25 - 20( \frac{\sqrt{5}}{\sqrt{2}} ) - a( \frac{\sqrt{5}}{\sqrt{2}}) +b =0.....[B]$

Adding Equation [A] and equation [B]

$50 +2b =0$

$b=-25$

Subtracting Equation [A] and equation [B]

$40(\sqrt{\frac{5}{2} } ) + 2a(\sqrt{\frac{5}{2} } ) =0$

$a = -20$

Therefore,

The value of $a = -20$

The value of $b=-25$

Keywords: equation, solution

Learn more about equation solution from brainly.com/question/12687078

#learnwithBrainly

3 0

3 years ago

In David’s class, the ratio of right-handed students to left-handed students is 24 to 6. How many right-handed students are ther

BaLLatris [955]

Answer:

There are 4 right handed students for each left hand because, you just divide 24 by 6 and you get 4 for each of the 6 left handed students.

Step-by-step explanation:

Or just group, by doing 6 circles and just put one in each of them so then you can see when all of the circles are taken up .

Thanks! Please Mark Brainiest! <3

~ BiologyisAmazing

6 0

3 years ago

Due soon pls help will give brianliest

Mamont248 [21]

I think it would be:

10, 12.5, 17, 19.5, 23

hope it helps

6 0

3 years ago

Read 2 more answers

HELP!!!!!!!!!!!!!!!!!!

padilas [110]

First, factor each expression by finding two terms whose product is equal to the c term and sum is equal to the b term.

x^2-x+6 --> (x-3)(x+2)

x^2-4 --> (x+2) (x-2)

Next, cancel the terms that are equal on the top and bottom.

You are left with the answer of: $\frac{x-3}{x-2}$

Hope this helps!!

8 0

3 years ago

Can the side lengths 12,15, and 13 form a triangle?​

Can the side lengths 12,15, and 13 form a triangle?