If each cube has edges 2 centimeters long, what is the volume of the blue-outlined prism?

Identify the hypothesis and conclusion of each conditional statement. Then write each sentence in if-then form

lukranit [14]

Answer:

If you place two collinear points, then they will fall on the same line

Step-by-step explanation:

?

3 0

3 years ago

Here is a triangular pyramid and its net.

galben [10]

Answer:

a) Area of the base of the pyramid = $15.6\ mm^{2}$

b) Area of one lateral face = $24\ mm^{2}$

c) Lateral Surface Area = $72\ mm^{2}$

d) Total Surface Area = $87.6\ mm^{2}$

Step-by-step explanation:

We are given the following dimensions of the triangular pyramid:

Side of triangular base = 6mm

Height of triangular base = 5.2mm

Base of lateral face (triangular) = 6mm

Height of lateral face (triangular) = 8mm

a) To find Area of base of pyramid:

We know that it is a triangular pyramid and the base is a equilateral triangle. $\text{Area of triangle = } \dfrac{1}{2} \times \text{Base} \times \text{Height} ..... (1)\\$

${\Rightarrow \text{Area of pyramid's base = }\dfrac{1}{2} \times 6 \times 5.2\\\Rightarrow 15.6\ mm^{2}$

b) To find area of one lateral surface:

Base = 6mm

Height = 8mm

Using equation (1) to find the area:

$\Rightarrow \dfrac{1}{2} \times 8 \times 6\\\Rightarrow 24\ mm^{2}$

c) To find the lateral surface area:

We know that there are 3 lateral surfaces with equal height and equal base.

Hence, their areas will also be same. So,

$\text{Lateral Surface Area = }3 \times \text{ Area of one lateral surface}\\\Rightarrow 3 \times 24 = 72 mm^{2}$

d) To find total surface area:

Total Surface area of the given triangular pyramid will be equal to <em>Lateral Surface Area + Area of base</em>

$\Rightarrow 72 + 15.6 \\\Rightarrow 87.6\ mm^{2}$

Hence,

a) Area of the base of the pyramid = $15.6\ mm^{2}$

b) Area of one lateral face = $24\ mm^{2}$

c) Lateral Surface Area = $72\ mm^{2}$

d) Total Surface Area = $87.6\ mm^{2}$

8 0

3 years ago

Please help, I didn't learn this in class!

NemiM [27]

Solve the following system using elimination:
{-2 x + 2 y + 3 z = 0 | (equation 1)
{-2 x - y + z = -3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)

Subtract equation 1 from equation 2:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x - 3 y - 2 z = -3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)

Multiply equation 2 by -1:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+3 y + 2 z = 3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)

Add equation 1 to equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+3 y + 2 z = 3 | (equation 2)
{0 x+5 y + 6 z = 5 | (equation 3)

Swap equation 2 with equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+3 y + 2 z = 3 | (equation 3)

Subtract 3/5 × (equation 2) from equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y - (8 z)/5 = 0 | (equation 3)

Multiply equation 3 by 5/8:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y - z = 0 | (equation 3)

Multiply equation 3 by -1:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)

Subtract 6 × (equation 3) from equation 2:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y+0 z = 5 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)

Divide equation 2 by 5:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)

Subtract 2 × (equation 2) from equation 1:
{-(2 x) + 0 y+3 z = -2 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
v0 x+0 y+z = 0 | (equation 3)

Subtract 3 × (equation 3) from equation 1:
{-(2 x)+0 y+0 z = -2 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)

Divide equation 1 by -2:
{x+0 y+0 z = 1 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)

Collect results:

Answer: {x = 1, y = 1, z = 0

7 0

4 years ago

Use the given graph. Determine the period of the function.

Ivahew [28]

Answer:

3

Step-by-step explanation:

Period of a function is the period after which the function attains the same value

in the graph attached with this problem we can see that

f(0)=1

the value of x for which function f(x) attains the value 1 again is at

x=3

f(3)=1

similarly , we see

f(6)=1 , f(9)=1

Hence we see that after every increased value of x by 3 units , we attain the same value of function . hence the period of the function is 3

6 0

3 years ago

Use an array and partial products to find

creativ13 [48]

Well done mate you’re correct.

3 0

3 years ago