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Mamont248 [21]
3 years ago
7

50 POINTS!!!!!WILL MARK BRAINLIEST IF YOU HELP GEOMETRY. LOOK AT PHOTO FOR QUESTIONS

Mathematics
1 answer:
ki77a [65]3 years ago
6 0

Answer:

hmm a chicken

Step-by-step explanation:

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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
4d^3-10 as <br> verbal expression
Rina8888 [55]
Four times d to the third minus ten
3 0
3 years ago
Rosa gets a 233 as a score out of 285. What is the grade, A, B, C, or F?
riadik2000 [5.3K]

Answer:

B

Step-by-step explanation:

233/285*100

0.81754*100

81.754%

3 0
3 years ago
What is the number of possible combinations of 14 objects taken 3 at a time?
maria [59]

Answer:

364.

Step-by-step explanation:

14C3 = 14! / 11! 3! which simplifies to:

14*13*12 /3*2*1

= 14*13* 2

= 364.

6 0
3 years ago
Date:
Nadusha1986 [10]

Answer:

#3   1/6, 1/6, 1/2, 1/3, 5/6.

#4   3/12, 4/12, 5/12, 8/12, 9/12

Step-by-step explanation:

#3 is numbers out of six (fractions).

#4 is the same thing.  You add all of the numbers (4, 3, and 5) to get the total and then you subtract to get the probability.

These are all I know.

7 0
3 years ago
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