All categories

3 years ago

A polyhedron has 6 vertices and 9 edges. How many faces does it have?

Mathematics

2 answers:

Sidana [21]3 years ago

7 0

5 faces
The polyhedron is a Triangular Prism.

arlik [135]3 years ago

3 0

Answer:

Step-by-step explanation:

You might be interested in

Draw three line segments

kodGreya [7K]

Here ya go!!! this should be correct

6 0

3 years ago

Suppose 3/4 is a root of a polynomial equation. what does this tell us about the leading coefficient and the constant term in th

snow_tiger [21]

Answer:

Nothing new about leading coefficient. Definition says it can't be zero. It can always be made equal to one by dividing through.

Constant term cannot be zero unless a root is zero.

Step-by-step explanation:

x-3/4 = 0 has root 3/4 constant term -3/4

(x-3/4)(x-b) = 0 has roots 3/4 and b, constant term 3b/4 which is zero only if b is zero.

(x-3/4)(x-b)(x-c) = 0 has roots 3/4, b, c, constant term -3bc/4 which is zero only if b or c is zero.

Etc. ...

5 0

3 years ago

A basketball has a volume of 23047 cm

Elodia [21]

Answer:

17.65693585 = r

Step-by-step explanation:

The volume of a sphere is given by

V = 4/3 pi r^3

23047 = 4/3 (3.14) r^3

23047 = 4.1866666666 r^3

Divide by 4.186666 on each side

23047/4.18666666 = 4.1866666666/4.18666666 *r^3

5504.856688 = r^3

Take the cubed root on each side

(5504.856688)^1/3 = r^3^(1/3)

17.65693585 = r

4 0

3 years ago

Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and

Lemur [1.5K]

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates $(x_A,y_A)$

Vectors AB and AB' are perpendicular, then

$\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0$

Vectors AC and AC' are perpendicular, then

$\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0$

Now, solve the system of two equations:

$\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.$

Subtract these two equations:

$5x_A-y_A-8=0\Rightarrow y_A=5x_A-8$

Substitute it into the first equation:

$x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2$

Then

$y_{A_{1,2}}=5\cdot \dfrac{34}{13}-8 \text{ or } 5\cdot 2-8\\ \\=\dfrac{66}{13}\text{ or } 2$

Rotation by 90° counterclockwise about A(2,2) gives image points B' and C' (see attached diagram)

8 0

3 years ago

(15) Geometry Help plz

MakcuM [25]

X = measure of angle 1
y = measure of angle 2
z = measure of angle 3
w = measure of angle 4

Focus on the bottom triangle. The three angles add to 180 degrees
(angle 2) + (angle 3) + 116 = 180
y+z+116 = 180
y+z= 180-116
y+z= 64

Since we have the bottom triangle as isosceles, this means that y = z, so
y+z = 64
y+y = 64
2y = 64
y = 64/2
y = 32
making z = 32 as well

Similarly, angle 1 and angle 4 are 32 degrees because the 116 angle is opposite the top left-most angle, and congruent to this angle. In other words, the bottom triangle is a mirror image of the top triangle.

The figure is a rhombus because all four sides are the same length (as shown by the tickmarks)

-----------------------------------------------------------

Answer:
This figure is a rhombus
All four angles (angle 1 through angle 4) are the same measure. They are each 32 degrees

6 0

3 years ago