All categories

2 years ago

A triangle has side lengths of 10, 11, and 15 . what type of triangle is it?

1 answer:

7 0

Answer: scalene and obtuse

Justification:

You can find the angles using the law of cosine:

c^2 = a^2 + b^2 - 2abcos(γ)

=> cos(γ) = [a^2 + b^2 - c^2] / (2ab)

1) cos(γ) = [10^2 + 11^2 - 15^2] / (2*10*11) = - 0.0181818

=> γ = arccos(-0.0181818} ≈ 91°

2) cos(α) = [b^2 + c^2 - a^2 ] / 2bc = [11^2 + 15^2 - 10^2] / (2*11*15] = 0.7454545

=> α = arccos(0.7454545) ≈ 41.8°

3) cos(β) = [a^2 + c^2 - b^2] / (2ac) = [10^2 + 15^2 - 11^2] /(2*10*15) = 0.68

=> β = arccos(0.68) ≈ 47.2°

4) Verification: 91° + 41.8° + 47.2° = 180°

5) The triangles with the three different sides are called scalenes (which you can tell with only the measures of the sides).

6) The triangles with one angle greater than 90° are called obtuse.

So, the triangle is scalene and obtuse.

You might be interested in

In Exercises 45–48, let f(x) = (x - 2)2 + 1. Match the function with its graph

MA_775_DIABLO [31]

Answer:

45) The function corresponds to graph A

46) The function corresponds to graph C

47) The function corresponds to graph B

48) The function corresponds to graph D

Step-by-step explanation:

We know that the function f(x) is:

$f(x)=(x-2)^{2}+1$

45)

The function g(x) is given by:

$g(x)=f(x-1)$

using f(x) we can find f(x-1)

$g(x)=((x-1)-2)^{2}+1=(x-3)^{2}+1$

If we take the derivative and equal to zero we will find the minimum value of the parabolla (x,y) and then find the correct graph.

$g(x)'=2(x-3)$

$2(x-3)=0$

$x=3$

Puting it on g(x) we will get y value.

$y=g(3)=(3-3)^{2}+1$

$y=g(3)=1$

Then, the minimum point of this function is (3,1) and it corresponds to (A)

46)

Let's use the same method here.

$g(x)=f(x+2)$

$g(x)=((x+2)-2)^{2}+1$

$g(x)=(x)^{2}+1$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2x$

$0=2x$

$x=0$

Evaluatinf g(x) at this value of x we have:

$g(0)=(x)^{2}+1$

$g(0)=1$

Then, the minimum point of this function is (0,1) and it corresponds to (C)

47)

Let's use the same method here.

$g(x)=f(x)+2$

$g(x)=(x-2)^{2}+1+2$

$g(x)=(x-2)^{2}+3$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2(x-2)$

$0=2(x-2)$

$x=2$

Evaluatinf g(x) at this value of x we have:

$g(2)=(2-2)^{2}+3$

$g(2)=3$

Then, the minimum point of this function is (2,3) and it corresponds to (B)

48)

Let's use the same method here.

$g(x)=f(x)-3$

$g(x)=(x-2)^{2}+1-3$

$g(x)=(x-2)^{2}-2$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2(x-2)$

$0=2(x-2)$

$x=2$

Evaluatinf g(x) at this value of x we have:

$g(2)=(2-2)^{2}-2$

$g(2)=-2$

Then, the minimum point of this function is (2,-2) and it corresponds to (D)

I hope it helps you!

8 0

2 years ago

Given the following sets.

asambeis [7]

A = {0, 1, 2, 3}

C = {0, a, 2, b}

A ∩ C = {0, 2} → E)

A set of the same elements from the set A and the set C.

4 0

3 years ago

Which area model can u use to fine * find 39 x73

sweet [91]

The area model you can use is D , cs 70 +3 & 30+9

3 0

2 years ago

For the functions f(x)=−7x^2−x and g(x)=9x^2−4x, find (f−g)(x) and (f−g)(1)

musickatia [10]

Answer:

See below.

Step-by-step explanation:

$f(x)=-7x^2-x\\g(x)=9x^2-4x\\(f-g)(x)=f(x)-g(x)\\(f-g)(x)=(-7x^2-x)-(9x^2-4x)\\(f-g)(x)=-7x^2-x-9x^2+4x\\(f-g)(x)=-16x^2+3x\\\\(f-g)(1)=-16(1)^2+3(1)\\(f-g)(1)=-16+3=-13$