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Volgvan
3 years ago
14

Angle efb is 80 what is the the size of angle x

Mathematics
1 answer:
quester [9]3 years ago
5 0

Answer: a) The size of angle x is 80^{\circ}.

b)  A. Corresponding angles.

Step-by-step explanation:

Since AD is parallel to EH and CG is a transversal intersecting them at B and F, then \angle ABF = \angle EFG  ...(i)

[Pair of corresponding angles are equal lie on the same side of the transversal]

Here, \angle EFG =80^{\circ} , \angle ABF = x

then, x =80^{\circ}  [Substitute values in (i)]

Hence, the size of angle x is 80^{\circ}.

Reason for this: Corresponding angles.

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Answer:

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Step-by-step explanation:

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This should be correct

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Choose a counterexample that proves that the conjecture below is false.. abc is a right triangle, so angle A measures 90 degrees
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Choose a counterexample that proves that the conjecture below is false.. abc is a right triangle, so angle A measures 90 degrees.

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I hope it helps, Regards.
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4 years ago
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3 years ago
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3 years ago
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Gravel is being dumped from a conveyor belt at a rate of 20 ft3 /min and its coarseness is such that it forms a pile in the shap
pantera1 [17]

Answer:

The height of the pile is increasing at the rate of  \mathbf{ \dfrac{20}{56.25 \pi}   \ \ \ \ \  ft/min}

Step-by-step explanation:

Given that :

Gravel is being dumped from a conveyor belt at a rate of 20 ft³ /min

i.e \dfrac{dV}{dt}= 20 \ ft^3/min

we know that radius r is always twice the   diameter d

i.e d = 2r

Given that :

the shape of a cone whose base diameter and height are always equal.

then d = h = 2r

h = 2r

r = h/2

The volume of a cone can be given by the formula:

V = \dfrac{\pi r^2 h}{3}

V = \dfrac{\pi (h/2)^2 h}{3}

V = \dfrac{1}{12} \pi h^3

V = \dfrac{ \pi h^3}{12}

Taking the differentiation of volume V with respect to time t; we have:

\dfrac{dV}{dt }= (\dfrac{d}{dh}(\dfrac{\pi h^3}{12})) \times \dfrac{dh}{dt}

\dfrac{dV}{dt }= (\dfrac{\pi h^2}{4} ) \times \dfrac{dh}{dt}

we know that:

\dfrac{dV}{dt}= 20 \ ft^3/min

So;we have:

20= (\dfrac{\pi (15)^2}{4} ) \times \dfrac{dh}{dt}

20= 56.25 \pi \times \dfrac{dh}{dt}

\mathbf{\dfrac{dh}{dt}= \dfrac{20}{56.25 \pi}   \ \ \ \ \  ft/min}

The height of the pile is increasing at the rate of  \mathbf{ \dfrac{20}{56.25 \pi}   \ \ \ \ \  ft/min}

8 0
3 years ago
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