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2 years ago

7

Can someone help me please

1 answer:

kifflom [539]2 years ago

4 0

9514 1404 393

Answer:

{Segments, Geometric mean}

{PS and QS, RS}

{PS and PQ, PR}

{PQ and QS, QR}

Step-by-step explanation:

The three geometric mean relationships are derived from the similarity of the triangles the similarity proportions can be written 3 ways, each giving rise to one of the geometric mean relations.

short leg : long leg = SP/RS = RS/SQ ⇒ RS² = SP·SQ

short leg : hypotenuse = RP/PQ = PS/RP ⇒ RP² = PS·PQ

long leg : hypotenuse = RQ/QP = QS/RQ ⇒ RQ² = QS·QP

I find it easier to remember when I think of it as <em>the segment from R is equal to the geometric mean of the two segments the other end is connected to</em>.

__

segments PS and QS, gm RS

segments PS and PQ, gm PR

segments PQ and QS, gm QR

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ioda

Answer: The first option with sss as part of the proof

Step-by-step explanation:

The lengths of the arcs drawn in the constructions are the same, so when the line is drawn, the lengths of the sides will all be the same. SSS is proof of congruence.

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7 0

2 years ago

Read 2 more answers

Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca

Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

$\frac{dQ}{dt} = -rt$

Which has the following solution:

$Q(t) = Q(0)e^{-rt}$

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that $Q(6) = (1-0.03)Q(0) = 0.97Q(0)$ . We use this to find r.

$Q(t) = Q(0)e^{-rt}$

$0.97Q(0) = Q(0)e^{-6r}$

$e^{-6r} = 0.97$

$\ln{e^{-6r}} = \ln{0.97}$

$-6r = \ln{0.97}$

$r = -\frac{\ln{0.97}}{6}$

$r = 0.0051$

So

$Q(t) = Q(0)e^{-0.0051t}$

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

$Q(t) = Q(0)e^{-0.0051t}$

$0.5Q(0) = Q(0)e^{-0.0051t}$

$e^{-0.0051t} = 0.5$

$\ln{e^{-0.0051t}} = \ln{0.5}$

$-0.0051t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.0051}$

$t = 135.9$

The half-life of the radioactive substance is 135.9 hours.

6 0

2 years ago

Help me thankyouuuuuuu

erica [24]

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2 years ago

Read 2 more answers

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skad [1K]

Answer: D

Step-by-step explanation:

Using the equation y = mx + b.

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Add y to both sides and the 0 can drop out.

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In this equation the m is equal to 1 and the b is equal to 0.

Taking this into account the answer is D.

8 0

2 years ago

How do you answer this

Sedbober [7]

you just need to multiply the length times width

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2 years ago