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Rom4ik [11]
2 years ago
11

The linear model represents the height, f(x), of a water balloon thrown off the roof of a building over time, x, measured in sec

onds:
A linear model with ordered pairs at 0, 60 and 2, 75 and 4, 75 and 6, 40 and 8, 20 and 10, 0 and 12, 0 and 14, 0. The x axis is labeled Time in seconds, and the y axis is labeled Height in feet.

Part A: During what interval(s) of the domain is the water balloon's height increasing? (2 points)

Part B: During what interval(s) of the domain is the water balloon's height staying the same? (2 points)

Part C: During what interval(s) of the domain is the water balloon's height decreasing the fastest? Use complete sentences to support your answer. (3 points)

Part D: Use the constraints of the real-world situation to predict the height of the water balloon at 16 seconds. Use complete sentences to support your answer. (3 points)

Mathematics
1 answer:
pashok25 [27]2 years ago
6 0

Answer:

Step-by-step explanation:

<h2><u>Part A</u></h2>

in interval ( 0 ; 2)

<h2><u>Part B</u></h2>

in interval (2; 4)

<h2><u>Part C</u></h2>

in interval  (4 ; 6 )

<h2><u>Part D</u></h2>

The graph shows that at first the ball rises up ; and then it is seen that it goes down and loses height to zero , from which it can be concluded that the height after 10 seconds remains unchanged and therefore the height of the ball after 16 seconds will be zero

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3^5 equals 243. Explain how to use that fact to quickly evaluate 3^6.
OlgaM077 [116]

Answer:

Simply multiply 243 with 3.

Step-by-step explanation:

3^6=729

243*3=729

4 0
3 years ago
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Will mark brainliest for the correct answer!
romanna [79]

Part (a)

Focus on triangle PSQ. We have

angle P = 52

side PQ = 6.8

side SQ = 5.4

Use of the law of sines to determine angle S

sin(S)/PQ = sin(P)/SQ

sin(S)/(6.8) = sin(52)/(5.4)

sin(S) = 6.8*sin(52)/(5.4)

sin(S) = 0.99230983787513

S = arcsin(0.99230983787513)

S = 82.889762826274

Which is approximate

------------

Use this to find angle Q. Again we're only focusing on triangle PSQ.

P+S+Q = 180

Q = 180-P-S

Q = 180-52-82.889762826274

Q = 45.110237173726

Which is also approximate.

A more specific name for this angle is angle PQS, which will be useful later in part (b).

------------

Now find the area of triangle PSQ

area of triangle = 0.5*(side1)*(side2)*sin(included angle)

area of triangle PSQ = 0.5*(PQ)*(SQ)*sin(angle Q)

area of triangle PSQ = 0.5*(6.8)*(5.4)*sin(45.110237173726)

area of triangle PSQ = 13.0074347717966

------------

Next we'll use the fact that RS:SP is 2:1.

This means RS is twice as long as SP. Consequently, this means the area of triangle RSQ is twice that of the area of triangle PSQ. It might help to rotate the diagram so that line PSR is horizontal and Q is above this horizontal line.

We found

area of triangle PSQ = 13.0074347717966

So,

area of triangle RSQ = 2*(area of triangle PSQ)

area of triangle RSQ = 2*13.0074347717966

area of triangle RSQ = 26.0148695435932

------------

We're onto the last step. Add up the smaller triangular areas we found

area of triangle PQR = (area of triangle PSQ)+(area of triangle RSQ)

area of triangle PQR = (13.0074347717966)+(26.0148695435932)

area of triangle PQR = 39.0223043153899

------------

<h3>Answer: 39.0223043153899</h3>

This value is approximate. Round however you need to.

===========================================

Part (b)

Focus on triangle PSQ. Let's find the length of PS.

We'll use the value of angle Q to determine this length.

We'll use the law of sines

sin(Q)/(PS) = sin(P)/(SQ)

sin(45.110237173726)/(PS) = sin(52)/(5.4)

5.4*sin(45.110237173726) = PS*sin(52)

PS = 5.4*sin(45.110237173726)/sin(52)

PS = 4.8549034284642

Because RS is twice as long as PS, we know that

RS = 2*PS = 2*4.8549034284642 = 9.7098068569284

So,

PR = RS+PS

PR = 9.7098068569284 + 4.8549034284642

PR = 14.5647102853927

-------------

Next we use the law of cosines to find RQ

Focus on triangle PQR

c^2 = a^2 + b^2 - 2ab*cos(C)

(RQ)^2 = (PR)^2 + (PQ)^2 - 2(PR)*(PQ)*cos(P)

(RQ)^2 = (14.5647102853927)^2 + (6.8)^2 - 2(14.5647102853927)*(6.8)*cos(52)

(RQ)^2 = 136.420523798282

RQ = sqrt(136.420523798282)

RQ = 11.6799196828694

--------------

We'll use the law of sines to find angle R of triangle PQR

sin(R)/PQ = sin(P)/RQ

sin(R)/6.8 = sin(52)/11.6799196828694

sin(R) = 6.8*sin(52)/11.6799196828694

sin(R) = 0.4587765387107

R = arcsin(0.4587765387107)

R = 27.3081879220073

--------------

This leads to

P+Q+R = 180

Q = 180-P-R

Q = 180-52-27.3081879220073

Q = 100.691812077992

This is the measure of angle PQR

subtract off angle PQS found back in part (a)

angle SQR = (anglePQR) - (anglePQS)

angle SQR = (100.691812077992) - (45.110237173726)

angle SQR = 55.581574904266

--------------

<h3>Answer: 55.581574904266</h3>

This value is approximate. Round however you need to.

8 0
3 years ago
Find the exact value of csc theta if tan theta = sqrt3 and the terminal side of theta is in Quadrant III.
WARRIOR [948]

Answer:

3rd option

Step-by-step explanation:

Using the identities

cot x = \frac{1}{tanx}

csc² x = 1 + cot² x

Given

tanθ = \sqrt{3} , then cotθ = \frac{1}{\sqrt{3} }

csc²θ = 1 + (\frac{1}{\sqrt{3} } )² = 1 + \frac{1}{3} = \frac{4}{3}

cscθ = ± \sqrt{\frac{4}{3} } = ± \frac{2}{\sqrt{3} }

Since θ is in 3rd quadrant, then cscθ < 0

cscθ = - \frac{2}{\sqrt{3} } × \frac{\sqrt{3} }{\sqrt{3} } = - \frac{2\sqrt{3} }{3}

8 0
3 years ago
Suri solved the equation x squared = 49 and found that x=7. What error did Suri make?
Anastaziya [24]

Answer:

x = \frac{+}{-} 7

Step-by-step explanation:

x^{2} = 49

\sqrt{49} can equal 7 or -7

7 x 7 = 49.  -7 x -7 = 49

7 0
3 years ago
I don't understand 1/2 +1/6 what the answer
romanna [79]

Answer:

4/6

Step-by-step explanation:

1/2=3/6+1/6=4/6

3 0
2 years ago
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