All categories

3 years ago

Classify the following triangle. Check all that apply

Mathematics

1 answer:

Kamila [148]3 years ago

5 0

Answer:

The triangle is obtuse, and isosceles.

Step-by-step explanation:

The triangle is obtuse because there is an angle above 90 degrees, which is 104 degrees. The triangle is also isosceles due to the two lines in the triangle which indicate that the size of each side are equal to each other. For a triangle to be isosceles, there have to be at least two sides equal to each other, and in this case the triangle has those two equal sides.

You might be interested in

Here is a tree diagram showing the sample space for two independent events. How many outcomes are there for the compound event?

insens350 [35]

The answer is 4 and I hope I helped foo

8 0

3 years ago

Co-ordinate Geometry

saul85 [17]

Answer:

The value of a = 14

Step-by-step explanation:

Given

(x₁, y₁) = (2, 4)

(x₂, y₂) = (6, a)

(x₃, y₃) = (-1, 1)

A = 9 sq.units

Area of a triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃) is:

$A=\frac{\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|}{2}$

substituting the values (x₁, y₁) = (2, 4), (x₂, y₂) = (6, a), (x₃, y₃) = (-1, 1), A = 9 in th formula

$A=\frac{\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|}{2}$

$9=\frac{\left|2\left(a-1\right)+6\left(1-4\right)+-1\left(4-a\right)\right|}{2}$

Multiply both sides by 2

$\frac{2\left|2\left(a-1\right)+6\left(1-4\right)-1\left(4-a\right)\right|}{2}=9\cdot \:2$

simplify

$\left|2\left(a-1\right)+6\left(1-4\right)-1\left(4-a\right)\right|=18$

As the area is always positive.

$2\left(a-1\right)+6\left(1-4\right)-1\cdot \left(4-a\right)=18$

$2\left(a-1\right)-18-\left(4-a\right)=18$

Add 18 to both sides

$2\left(a-1\right)-18-\left(4-a\right)+18=18+18$

simplify

$2\left(a-1\right)-\left(4-a\right)=36$

$3a-6=36$

$3a=42$

Divide both sides by 3

$\frac{3a}{3}=\frac{42}{3}$

$a=14$

Thus, the value of a = 14

7 0

2 years ago

Find the flux of F=(x^5+y^5+z^5-2x-3y-4z)i+sin(2y)j+4zsin^2(y)k across the surface of the tetrahedron bounded by the coordinate

Tju [1.3M]

Use the divergence theorem.

$\mathbf F(x,y,z)=(x^5+y^5+z^5-2x-3y-4z)\,\mathbf i+\sin2y\,\mathbf j+4z\sin^2y\,\mathbf k$
$\implies(\nabla\cdot\mathbf F)(x,y,z)=\dfrac{\partial(x^5+y^5+z^5-2x-3y-4z)}{\partial x}+\dfrac{\partial(\sin2y)}{\partial y}+\dfrac{\partial(4z\sin^2y)}{\partial z}=5x^4-2+2\cos2y+4\sin^2y$

The flux of $\mathbf F$ across the tetrahedron's surface $S$ is then given by the integral of $\nabla\cdot\mathbf F$ over the interior of the tetrahedron $\mathbf R$ .

$\displaystyle\iint_S\mathbf F\cdot\mathrm dS=\iiint_T\nabla\cdot\mathbf F\,\mathrm dV$
$=\displaystyle\int_{x=0}^{x=1}\int_{y=0}^{y=1-x}\int_{z=0}^{z=1-x-y}(5x^4-2+2\cos2y+4\sin^2y)\,\mathrm dz\,\mathrm dy\,\mathrm dx$
$=\dfrac1{42}$