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agasfer [191]
3 years ago
6

Which solid has one face, zero lateral faces, one base, one vertex, and zero edges?

Mathematics
1 answer:
Ratling [72]3 years ago
7 0
The correct answer is cone.

There's only one face, and only one vertex that leads to a single base. There are no edges because the base perimeter is not considered to be an edge because it's a vertex.
You might be interested in
8) How many packs of DVD's can you buy with 77 dollars<br> if one pack costs 11 dollars ?
Vladimir79 [104]
You can buy 7 packs, because if you multiply 11x7 you would get 77. hope that helps
5 0
3 years ago
In right ABC, AN is the altitude to the hypotenuse. FindBN, AN, and AC,if AB =2 5 in, and NC= 1 in.
Rama09 [41]

From the statement of the problem, we have:

• a right triangle △ABC,

,

• the altitude to the hypotenuse is denoted AN,

,

• AB = 2√5 in,

,

• NC = 1 in.

Using the data above, we draw the following diagram:

We must compute BN, AN and AC.

To solve this problem, we will use Pitagoras Theorem, which states that:

h^2=a^2+b^2\text{.}

Where h is the hypotenuse, a and b the sides of a right triangle.

(I) From the picture, we see that we have two sub right triangles:

1) △ANC with sides:

• h = AC,

,

• a = ,NC = 1,,

,

• b = NA.

2) △ANB with sides:

• h = ,AB = 2√5,,

,

• a = BN,

,

• b = NA,

Replacing the data of the triangles in Pitagoras, Theorem, we get the following equations:

\begin{cases}AC^2=1^2+NA^2, \\ (2\sqrt[]{5})^2=BN^2+NA^2\text{.}\end{cases}\Rightarrow\begin{cases}NA^2=AC^2-1, \\ NA^2=20-BN^2\text{.}\end{cases}

Equalling the last two equations, we have:

\begin{gathered} AC^2-1=20-BN^2.^{} \\ AC^2=21-BN^2\text{.} \end{gathered}

(II) To find the values of AC and BN we need another equation. We find that equation applying the Pigatoras Theorem to the sides of the bigger right triangle:

3) △ABC has sides:

• h = BC = ,BN + 1,,

,

• a = AC,

,

• b = ,AB = 2√5,,

Replacing these data in Pitagoras Theorem, we have:

\begin{gathered} \mleft(BN+1\mright)^2=(2\sqrt[]{5})^2+AC^2 \\ (BN+1)^2=20+AC^2, \\ AC^2=(BN+1)^2-20. \end{gathered}

Equalling the last equation to the one from (I), we have:

\begin{gathered} 21-BN^2=(BN+1)^2-20, \\ 21-BN^2=BN^2+2BN+1-20 \\ 2BN^2+2BN-40=0, \\ BN^2+BN-20=0. \end{gathered}

(III) Solving for BN the last quadratic equation, we get two values:

\begin{gathered} BN=4, \\ BN=-5. \end{gathered}

Because BN is a length, we must discard the negative value. So we have:

BN=4.

Replacing this value in the equation for AC, we get:

\begin{gathered} AC^2=21-4^2, \\ AC^2=5, \\ AC=\sqrt[]{5}. \end{gathered}

Finally, replacing the value of AC in the equation of NA, we get:

\begin{gathered} NA^2=(\sqrt[]{5})^2-1, \\ NA^2=5-1, \\ NA=\sqrt[]{4}, \\ AN=NA=2. \end{gathered}

Answers

The lengths of the sides are:

• BN = 4 in,

,

• AN = 2 in,

,

• AC = √5 in.

7 0
1 year ago
The course grade in a statistics class is the average of the scores on five examinations. Suppose that a student's scores on the
nadya68 [22]

Answer:

84 is the highest possible course average

Step-by-step explanation:

Total number of examinations = 5

Average = sum of scores in each examination/total number of examinations

Let the score for the last examination be x.

Average = (66+78+94+83+x)/5 = y

5y = 321+x

x = 5y -321

If y = 6, x = 5×6 -321 =-291.the student cannot score -291

If y = 80, x = 5×80 -321 =79.he can still score higher

If If y = 84, x = 5×84 -321 =99.This would be the highest possible course average after the last examination.

If y= 100

The average cannot be 100 as student cannot score 179(maximum score is 100)

8 0
3 years ago
How many solutions does the following system of equations have?<br> 2y= 5x + 4<br> 1 = 3x + 2
Elden [556K]

Looks like a pretty regular system, no parallel lines.  In fact the second equation is a vertical line as written; not sure if that's a typo.  Either way,

Answer: EXACTLY ONE SOLUTION

4 0
3 years ago
N-
pogonyaev

Answer:

The answer is (0, 2)

Step-by-step explanation:

Ive also had this question and viewed iver it and said it was right this was some time ago but im glad to help :)

6 0
3 years ago
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