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3 years ago

Can someone give me the answers asap! 15 points

Mathematics

1 answer:

Eduardwww [97]3 years ago

3 0

In the first question,
Since D is the midpoint of line segment AC.
therefore, AD=DC.

In triangles, ABD & CBD.
Since, AB is congruent to CB and BD is a common side between those two triangles.
Therefore, AD is congruent to DC.

Finally, in triangles DEA & DEC
Since, ED line segment is a common side between those triangles and the measure of angle EDA is as the same of angle EDC which are equal to 90 degrees and AD is congruent to DC.
Therefore, triangle EDA is congruent to triangle EDC ( you have to take care of the triangle order letters in congruence) and EC is congruent to EA.

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The altitude of a triangle is increasing at a rate of 1.500 centimeters/minute while the area of the triangle is increasing at a

puteri [66]

Answer:

The base of the triangle is shrinking at a rate of $\frac{131}{32}$ centimeters per minute.

Step-by-step explanation:

The formula of the area of a triangle is given by the following expression:

$A = \frac{1}{2}\cdot b \cdot h$

Where:

$A$ - Area of the triangle, measured in square centimeters.

$b$ - Base of the triangle, measured in centimeters.

$h$ - Height of the triangle, measured in centimeters.

The base of the triangle is:

$b = \frac{2\cdot A}{h}$

If $A = 98000\,cm^{2}$ and $h = 8000\,cm$ , the base of the triangle is:

$b = \frac{2\cdot (98000\,cm^{2})}{8000\,cm}$

$b = 24.5\,cm$

The rate of change of the area of the triangle in time, measured in minutes, is obtained after differentiating by rule of chain and using deriving rules:

$\frac{dA}{dt} = \frac{1}{2}\cdot h\cdot \frac{db}{dt} + \frac{1}{2}\cdot b \cdot \frac{dh}{dt}$

$\frac{dA}{dt} = \frac{1}{2} \cdot \left(h\cdot \frac{db}{dt}+b \cdot \frac{dh}{dt} \right)$

The rate of change of the base of the triangle is now cleared:

$2\cdot \frac{dA}{dt} = h\cdot \frac{db}{dt} + b\cdot \frac{dh}{dt}$

$h\cdot \frac{db}{dt} = 2\cdot \frac{dA}{dt}-b\cdot \frac{dh}{dt}$

$\frac{db}{dt} = \frac{2\cdot \frac{dA}{dt} - b \cdot \frac{dh}{dt} }{h}$

Given that $\frac{dA}{dt} = 2000\,\frac{cm^{2}}{min}$ , $b = 24.5\,cm$ , $\frac{dh}{dt} = 1500\,\frac{cm}{min}$ and $h = 8000\,cm$ , the rate of change of the base of the triangle is:

$\frac{db}{dt} = \frac{2\cdot \left(2000\,\frac{cm^{2}}{min} \right)-(24.5\,cm)\cdot \left(1500\,\frac{cm}{min} \right)}{8000\,cm}$

$\frac{db}{dt} = -\frac{131}{32}\,\frac{cm}{min}$

The base of the triangle is shrinking at a rate of $\frac{131}{32}$ centimeters per minute.

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3 years ago

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zlopas [31]

Answer:

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The data provided is:

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The histogram of the data is attached below.

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