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2 months ago

True or False? A circle could be circumscribed about the quadrilateral below.

Mathematics

2 answers:

Svet_ta [12.7K]2 months ago

7 0

Answer:

The answer is: False

Step-by-step explanation:

If the sum of opposite angles in a quadrilateral equals 180°, then it is possible to draw a circumscribed circle around the quadrilateral.

In this case, $\angle B+\angle D= 110\°+90\° = 180\°$

but, $\angle A+\angle C= 70\°+90\°= 160\°$

Consequently, a circle cannot be circumscribed around this quadrilateral.

PIT_PIT [12.4K]2 months ago

4 0

Answer:

the answer is false

Step-by-step explanation:

the apex indicated this

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1 month ago

The mass of a colony of bacteria, in grams, is modeled by the function P given by P(t)=2+5tan^−1.(t/2), where t is measured in d

AnnZ [12381]

Answer:

1.21 g/day

Step-by-step explanation:

We start with the fact that

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$P(t)=2+5tan^{-1}(\frac{t}{2})$

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Next, we differentiate with respect to t

$P'(t)=5(\frac{1}{1+\frac{t^2}{4}}\times \frac{1}{2})$

Using the formula $\frac{d(tan^{-1}(x)}{dx}=\frac{1}{1+x^2}$

$P'(t)=\frac{5}{2}(\frac{4}{4+t^2})$

$P'(t)=\frac{10}{4+t^2}$

We know that P(t)=6

Now, substitute this value

$6=2+5tan^{-1}(\frac{t}{2})$

$5tan^{-1}(\frac{t}{2})=6-2=4$

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$\frac{t}{2}=tan(\frac{4}{5})$

$t=2tan(\frac{4}{5})$

Insert the given value of t

$P'(2tan\frac{4}{5})=\frac{10}{4+4tan^2(\frac{4}{5})}$

$P'(2tan\frac{4}{5})=\frac{10}{4}\times \frac{1}{1+tan^2(\frac{4}{5})}$

We understand that $1+tan^2\theta=sec^2\theta$

Applying the formula

$P'(2tan(\frac{4}{5})=\frac{5}{2}\times \frac{1}{sec^2(\frac{4}{5})}$

$P'(2tan\frac{4}{5})=\frac{5}{2}\times cos^2(\frac{4}{5})$

By employing $cos^2x=\frac{1}{sec^2x}$

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Consequently, the instantaneous rate at which the mass of the colony changes is=1.21g/day

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3 months ago

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