1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Otrada [13]
3 years ago
6

A hovercraft takes off from a platform. It’s height (in meters), x seconds is modeled by h(x)=-(x-11)(x+3). What is the height o

f the hovercraft at the time of the takeoff?
Mathematics
1 answer:
Eddi Din [679]3 years ago
6 0

Answer:

The height of the hovercraft at the time of takeoff is  33  meters.

Step-by-step explanation:

Height of hovercraft  x  seconds after takeoff  =  h(x)  =  -(x - 11)(x + 3)

Height after takeoff is  =  h(0)  =  -(0 - 11)(0 + 3)  =  -(-11)(3)  =  33

So the correct answer is 33 meters. The height of the hover craft at the time of the takeoff is 33 metres.

Answer: 33  meters.

<em><u></u></em>

<em><u>Please mark brainliest</u></em>

<em><u>Hope this helps</u></em>

You might be interested in
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
Find the domain of f(x)=x-9/x squared-81
vagabundo [1.1K]

f(x)=\dfrac{x-9}{x^2-81}\\\\\text{The domain:}\\\\x^2-81\neq0\qquad\text{add 81 to both sides}\\\\x^2\neq81\to x\neq\pm\sqrt{81}\\\\x\neq-9\ \wedge\ x\neq9\\\\Answer:\ \boxed{\text{All real numbers except -9 and 9}}\to\boxed{x\in\mathbb{R}-\{-9,\ 9\}}

7 0
3 years ago
Proportions in Triangles (2)
forsale [732]

Answer:

  x = 6

Step-by-step explanation:

An angle bisector divides the segments on either side of it so they are proportional. That is ...

  x/12 = 5/10

  x = 12(5/10) = 6 . . . . . multiply by 12

6 0
3 years ago
Which of the following are measurements of the sides of a right triangle?
Anna [14]

the answer is D. all of the above

6 0
3 years ago
Claire jogs 6 miles per hour and Megan jogs 5 miles per hour. they start together at their campsite and jog to an outpost 18 mil
horrorfan [7]

Answer:

16.364 miles

Step-by-step explanation:

Speed of Claire = 6 miles per hour

Speed of Megan= 5 miles per hour

Distance from camp site to out post = 18 miles

Time for Claire to travel=18/6= 3 hours

Time for Megan to travel = 18/5 = 3.6 hours

The time they meet will be equal

Claire distance= 18 + x

Megan distance = 18-x

Claire time =( 18+x)/6

Megan Tim =( 18-x)/5

(18+x)/6 =(18-x)/5

5(18+x)= 6(18-x)

90 +5x = 108-6x

11x=18

X=18/11

X=1.636 Miles

So the distance from the Outpost to where they meet = 18-1.636

= 16.364 miles

6 0
3 years ago
Other questions:
  • What is the product of 5.86 × 10–7 and 3.1 × 104 1. Write the expression: (5.86 × 10–7)(3.1 × 104) 2. Rearrange the expression:
    13·2 answers
  • (x1/6)3 simplify the expression
    13·2 answers
  • Blueprint paper can be purchased in a roll that is 36 inches wide. If unrolled, the paper would extend 150 feet in length. What
    13·1 answer
  • A $1,000 face value bond is currently quoted at 100.8. the bond pays semiannual payments of $22.50 each and matures in six years
    15·1 answer
  • The bicycle is 5 feet long. How many inches is that?
    7·1 answer
  • What does r equal? <br>5(3-5r) + 3r = -(5r + 2)
    13·1 answer
  • Dale mixed 5.9 grams of salt into a pot of soup he was cooking. Before he served the soup, Dale added 0.31 grams of salt. How mu
    9·1 answer
  • Three angles are in the ratio 2 : 3 : 5 The smallest angle is 50 degrees Work out the sizes of the other two angles
    8·1 answer
  • Need this fast yal. im gonna give brainly
    13·2 answers
  • Answers pls!!!<br><br><br> math geniuses help!
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!