Can you guys please help me with questions 1-12 I’ll appreciate it and if you do help me can you SHOW YOUR WORK please.

How many unique planes can be determined by four noncoplanar points?

marta [7]

Answer:

Four unique planes

Step-by-step explanation:

Given that the points are non co-planar, triangular planes can be formed by the joining of three points

The points will therefore appear to be at the corners of a triangular pyramid or tetrahedron such that together the four points will form a three dimensional figure bounded by triangular planes

The number of triangular planes that can therefore be formed is given by the combination of four objects taking three at a time as follows;

₄C₃ = 4!/(3!×(4-3)! = 4

Which gives four possible unique planes.

3 0

3 years ago

The point (2,-5) is reflected over the y-axis. Which of the following points is the image?

docker41 [41]

2,5 would be the answer

7 0

3 years ago

Complete the equation describing how x and y are related.

Sav [38]

Answer: The "?" would equal 1.

Step-by-step explanation: This is simply a line with a slope of 1 and a y-intercept of zero.

5 0

2 years ago

8 freshmen, 9 sophomores, 9 juniors, and 7 seniors are eligible to be on a committee.

loris [4]

Answer: 11113200

Step-by-step explanation:

We know that , the number of combination of choosing r things from n things is given by :-

$^nC_r=\dfrac{n!}{r!(n-r)!}$

Then , the number of ways to choose a dancing committee if it is to consist of 4 freshmen, 5 sophomores, 2 juniors, and 3 seniors :-

$^8C_4\times ^9C_5\times^9C_2\times^7C_3$

$=\dfrac{8!}{(8-4)!4!}\times\dfrac{9!}{(9-5)!5!}\times\dfrac{9!}{(9-2)!2!}\times\dfrac{7!}{3!(7-3)!}\\\\ =\dfrac{8\times7\times6\times5\times4!}{4!4!}\times\dfrac{9\times8\times7\times6\times5!}{4!5!}\times\dfrac{9\times8\times7!}{7!2!}\times\dfrac{7\times6\times5\times4!}{3!4!}\\\\=11113200$

∴ The number of ways to choose a dancing committee if it is to consist of 4 freshmen, 5 sophomores, 2 juniors, and 3 seniors = 11113200

6 0

3 years ago

If ∠BAC = 17° and ∠CED = 17° are the two triangles, ΔBAC and ΔCED similar? If so, by what criterion?

Alborosie

No, not possible to tell that the the two triangles, ΔBAC and ΔCED

are similar ⇒ answer D

Step-by-step explanation:

Let us revise the cases of similarity

1. AAA similarity : two triangles are similar if all three angles in the first

triangle equal the corresponding angle in the second triangle

2. AA similarity : If two angles of one triangle are equal to the

corresponding angles of the other triangle, then the two triangles

are similar.

3. SSS similarity : If the corresponding sides of the two triangles are

proportional, then the two triangles are similar.

4. SAS similarity : In two triangles, if two sets of corresponding sides

are proportional and the included angles are equal then the two

triangles are similar.

In the two triangles BAC and CED

∵ m∠BAC = 17°

∵ m∠CED = 17°

∴ m∠BAC = m∠CED

But we need another pair of angles to prove that the two triangles are

similar by AA similarity criterion

<em>OR </em>

The lengths of sides BA , CA and CE , DE to show that

$\frac{BA}{CE}=\frac{CA}{DE}$ = constant ratio and prove that the

two triangles are similar by SAS similarity criterion

So it is not possible to prove that the two triangles are similar

No, not possible to tell that the the two triangles, ΔBAC and ΔCED

are similar

Learn more:

You can learn more about triangles in brainly.com/question/4354581

#LearnwithBrainly

5 0

3 years ago

Read 2 more answers