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DENIUS [597]
3 years ago
15

Find the area of the triangle whose base is 16 cm and height is 7 centimetre​

Mathematics
2 answers:
svetoff [14.1K]3 years ago
4 0

Answer:

56 cm²

Step-by-step explanation:

The area (A) of a triangle is calculated as

A = \frac{1}{2} bh ( b is the base and h the height )

Here b = 16 and h = 7 , then

A = \frac{1}{2} × 16 × 7 = 8 × 7 = 56 cm²

astra-53 [7]3 years ago
3 0

Answer:

56

Step-by-step explanation:

all you have to do is use the triangle area formula what is 1/2 base times height so the entire equation would be 1/2× 16×7=56

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t = \frac{A-P}{Pr}

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23,802 divided by 34 <br> HALP
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700.058824 but rounded to the nearest tenth it would be 700.1.

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6 0
3 years ago
g A plane flying horizontally at an altitude of 2 miles and a speed of 550 mi/h passes directly over a radar station. Find the r
guapka [62]

Answer:

The rate at which the distance from the plane to the station is increasing is when it is 5 miles away from the station is 504 mi/h

Step-by-step explanation:

An illustrative diagram for the scenario is shown in the attachment below.

In the diagram, y represent the altitude, z is the horizontal distance from the plane to the station and x is the distance from the plane to the station.

From the Pythagorean theorem, we can write that

x² = y² + z²

Differentiate this with respect to time t

That is,

\frac{d}{dt}(x^{2}) = \frac{d}{dt}(y^{2}) + \frac{d}{dt}(z^{2})

= 2x(\frac{dx}{dt}) = 2y(\frac{dy}{dt}) + 2z(\frac{dz}{dt})

x(\frac{dx}{dt}) = y(\frac{dy}{dt}) + z(\frac{dz}{dt})

The rate at which the distance from the plane to the station is increasing is \frac{dx}{dt}

\frac{dy}{dt} is the rate at which the altitude is increasing, since the altitude is 2, that is constant, \frac{dy}{dt} = 0.

\frac{dz}{dt} is the rate at which the horizontal distance is increasing which is the speed, that is, \frac{dz}{dt} = 550 mi/h

y = 2 and x = 5

Now, we will determine z when x = 5.

From x² = y² + z²

5² = 2² + z²

25 = 4 + z²

z² = 25-4

z² = 21

z =√21

Putting all the values into the equation

x(\frac{dx}{dt}) = y(\frac{dy}{dt}) + z(\frac{dz}{dt})

5(\frac{dx}{dt}) = 2(0) + \sqrt{21} (550)

5(\frac{dx}{dt}) = \sqrt{21} (550)

\frac{dx}{dt} = \frac{\sqrt{21} (550)}{5}

\frac{dx}{dt} = 504 mi/h

Hence, the rate at which the distance from the plane to the station is increasing is when it is 5 miles away from the station is 504 mi/h.

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