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bearhunter [10]
3 years ago
13

I Need help with 4-7

Mathematics
1 answer:
Murrr4er [49]3 years ago
4 0

Answer:

Step-by-step explanation:

4) parallel because 118° is a supplement to 62° and the corresponding angles are both 118°

5) NOT parallel. The labeled angles sum to 120° and would sum to 180° for parallel lines.

6) NOT parallel. see pic.

If parallel, extending a line to intersect ℓ₁ makes an opposite internal angle which would also be 48°. The created triangle would have its third angle at 180 - 90 - 48 = 42° which is opposite a labeled 48° angle, which is false, so the lines cannot be parallel

7)

b = 78°  as it corresponds with a labeled angle above it

a = 180 - 78 = 102° as angles along a line from a common vertex sum to 180

f = is an opposite angle to 180 - 78 - 44 = 58° as angles along a line from a common vertex sum to 180

e = 180 - 90 - 64 = 26° as angles along a line from a common vertex sum to 180

c = 58° as it corresponds with f

d = 180 - 58 = 122° as angles along a line from a common vertex sum to 180

 

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the height h(t) of a trianle is increasing at 2.5 cm/min, while it's area A(t) is also increasing at 4.7 cm2/min. at what rate i
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Answer:

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

Step-by-step explanation:

From Geometry we understand that area of triangle is determined by the following expression:

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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.

By Differential Calculus we deduce an expression for the rate of change of the area in time:

\frac{dA}{dt} = \frac{1}{2}\cdot \frac{db}{dt}\cdot h + \frac{1}{2}\cdot b \cdot \frac{dh}{dt} (Eq. 2)

Where:

\frac{dA}{dt} - Rate of change of area in time, measured in square centimeters per minute.

\frac{db}{dt} - Rate of change of base in time, measured in centimeters per minute.

\frac{dh}{dt} - Rate of change of height in time, measured in centimeters per minute.

Now we clear the rate of change of base in time within (Eq, 2):

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

\frac{db}{dt} = \frac{2}{h}\cdot \frac{dA}{dt} -\frac{b}{h}\cdot \frac{dh}{dt} (Eq. 3)

The base of the triangle can be found clearing respective variable within (Eq. 1):

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

If we know that A = 130\,cm^{2}, h = 15\,cm, \frac{dh}{dt} = 2.5\,\frac{cm}{min} and \frac{dA}{dt} = 4.7\,\frac{cm^{2}}{min}, the rate of change of the base of the triangle in time is:

b = \frac{2\cdot (130\,cm^{2})}{15\,cm}

b = 17.333\,cm

\frac{db}{dt} = \left(\frac{2}{15\,cm}\right)\cdot \left(4.7\,\frac{cm^{2}}{min} \right) -\left(\frac{17.333\,cm}{15\,cm} \right)\cdot \left(2.5\,\frac{cm}{min} \right)

\frac{db}{dt} = -2.262\,\frac{cm}{min}

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

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Answer:

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