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4 years ago

DOES ANYONE KNOW THIS?????

Mathematics

2 answers:

storchak [24]4 years ago

8 0

Answer:

Step-by-step explanation:

alternative angle theorem lets you know that angle acb is equal to cae and you can find acb since you know all angles add up to 180. 180-105-47=28

dedylja [7]4 years ago

5 0

Answer:

if i;m right its 45 degrees

Step-by-step explanation:

You might be interested in

Simplify the equation given in the image

IRINA_888 [86]

Answer:

b^20/a^12

Step-by-step explanation:

b to the power of 20 over a to the power of 12

6 0

3 years ago

Given the set of vertices, determine whether parallelogram ABCD is a rhombus, rectangle or square. List all that apply. A(7,-4),

Sloan [31]

Given:

Vertices of a parallelogram ABCD are A(7,-4), B(-1,-4), C(-1,-12), D(7, -12).

To find:

Whether the parallelogram ABCD is a rhombus, rectangle or square.

Solution:

Distance formula:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using distance formula, we get

$AB=\sqrt{(-4-(-4))^2+(-1-7)^2}$

$AB=\sqrt{(-4+4)^2+(-8)^2}$

$AB=\sqrt{0+64}$

$AB=8$

Similarly,

$BC=\sqrt{(-1-(-1))^2+(12-(-4))^2}=8$

$CD=\sqrt{(7-(-1))^2+(-12-(-12))^2}=8$

$AD=\sqrt{(7-7)^2+(-12-(-4))^2}=8$

All sides of parallelogram are equal.

$AC=\sqrt{(-1-7)^2+(-12-(-4))^2}=8\sqrt{2}$

$BD=\sqrt{(7-(-1))^2+(-12-(-4))^2}=8\sqrt{2}$

Both diagonals are equal.

Since, all sides are equal and both diagonals are equal, therefore, the parallelogram ABCD is a square.

We know that, a square is special case of rectangles and rhombus.

So, parallelogram ABCD is a rhombus, rectangle or square. Therefore, the correct option is c.

7 0

3 years ago

Simplify 2^2[4+(3-6)^2]-13

HACTEHA [7]

Answer:

It's 19.

Step-by-step explanation:

Use order of operations (PEMDAS):

2^2[4+(3-6)^2]-13

Parentheses first:

= 2^2[4 + (-2)^2] - 13

= 2^2 ( 4 + 4) - 13

= 2^2 * 8 - 13

Now work out the exponent:

= 4*8 - 13

Now the multiplication:

= 32 - 13

= 19.

8 0

3 years ago

What is the value of x in the equation 3x- 4y =65, when y= 4?

Alexus [3.1K]

3x-4y=65

3x=65+4y

x=(65+4y)/3, when y=4

x=(65+4*4)/3

x=(65+16)/3

x=81/3

x=27

3 0

3 years ago

Read 2 more answers

Write a coordinate proof to prove that the segment that joins the vertex angle of an isosceles triangle to the midpoint of its b

Natalija [7]

Answer:

Slope of the base segment line is zero, hence base segment is horizontal

Slope of the segment that joins the vertex angle to the midpoint of its base is undefined hence the line is vertical

Therefore, angle between base segment line and the segment line from the vertex angle to the midpoint is perpendicular

Therefore, the segment that joins the vertex angle to an isosceles triangle to the midpoint of its base is perpendicular to the base

Step-by-step explanation:

Here we prove the required relation as follows;

Let the isosceles be ABC

The coordinates of the points are

C = (0, 0) (Vertex)

A = (-a, b)

C = (a, b)

P = (0, b) (Midpoint of base)

Therefore, the gradient or slope of the base AC is presented as follows;

$Slope, m \, of \. base, \, AC= \frac{dy}{dx} = \frac{b - b}{a - (-a)} = \frac{0}{2\cdot a} = 0$

Hence, segment AC is vertical

$Slope, m \, of \, segment \ CP \, joining \, vertex, \, to \, midpoint, P, on \,AC = \frac{0 - b}{0 - 0} = \frac{-b}{0} = Undifined.$

Hence, segment CP is vertical

Therefore, the segment that joins the vertex angle to an isosceles triangle to the midpoint of its base is perpendicular to the base.

6 0

3 years ago