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sergeinik [125]
3 years ago
5

The height of the sail on a boat is 7 feet less than 3 times the length of its base. If the The area of the sail is 68 square fe

et, find its height and the length of the base.
Mathematics
1 answer:
Dima020 [189]3 years ago
3 0

Step-by-step explanation:

It is given that,

The height of the sail on a boat is 7 feet less than 3 times the length of its base.

Let the length of the base is x.

ATQ,

Height = (3x-7)

Area of the sail is 68 square feet.

Formula for area is given by :

A=lb\\\\68=x(3x-7)\\\\3x^2-7x=68\\\\3x^2-7x-68=0

x = 8 feet and x = -3.73 feet

So, length is 8 feet

Height is 3(8)-7 = 17 feet.

So, its height and the length of the base is 17 feet and 8 feet respectively.

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L=length\\
W=width\\
--------\\
P=84\\
L=\stackrel{\textit{twice the width}}{2W}
\end{cases}
\\\\\\
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5 0
4 years ago
Which expression is equivalent to<br> (6x-9x) - (2x - 3)?
NikAS [45]

Answer:

6x^2 -9x  -2x-3

6x^2 -11x +3

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
A person stands 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 meters
In-s [12.5K]

Answer:

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

Step-by-step explanation:

Given that,

A person stand 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 m/s.

From Pythagorean Theorem,

(The distance between car and person)²= (The distance of the car from intersection)²+ (The distance of the person from intersection)²+

Assume that the distance of the car from the intersection and from the person be x and y at any time t respectively.

∴y²= x²+10²

\Rightarrow y=\sqrt{x^2+100}

Differentiating with respect to t

\frac{dy}{dt}=\frac{1}{2\sqrt{x^2+100}}. 2x\frac{dx}{dt}

\Rightarrow \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}. \frac{dx}{dt}

Since the car driving towards the intersection at 13 m/s.

so,\frac{dx}{dt}=-13

\therefore \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}.(-13)

Now

\therefore \frac{dy}{dt}|_{x=24}=\frac{24}{\sqrt{24^2+100}}.(-13)

               =\frac{24\times (-13)}{\sqrt{676}}

               =\frac{24\times (-13)}{26}

               = -12 m/s

Negative sign denotes the distance between the car and the person decrease.

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

8 0
3 years ago
What is the distance between the points H(-5,9) and J(1,6)?
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Answer:The answer would be b

Step-by-step explanation:

8 0
3 years ago
Which equation is graphed in the figure?
djyliett [7]

Answer:

so the answer would be A. 7/5x +2

if you try graphing all the other equations it would be either over the picture or just not in the right place.

5 0
3 years ago
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