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3 years ago

Which pair of undefined terms is used to define the term parallel lines?

Mathematics

2 answers:

LuckyWell [14K]3 years ago

8 0

Answer:

Point and line

belka [17]3 years ago

6 0

Answer:

The answer is A: point and line : )

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Emanuel surveyed a random sample of 50 subscribers to Auto Wheel magazine about the number of cars that they own. Of the subscri

Nadusha1986 [10]

(Choice B) B 84 I did this a while ago

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3 years ago

Read 2 more answers

कम्प्युटरको गति मापन गर्ने एकाइहरूलाई वर्णन गर्नुहोस् ।

Anit [1.1K]

Answer:

मापनको एकाईलाई हर्ट्ज (हर्ट्ज) भनिन्छ, जुन प्रति सेकंड टेक्निकली एक चक्र हो, घडीको गति मापन गर्न प्रयोग हुन्छ। ... कम्प्युटरको घडी गति सामान्यतया मेगाहेर्ट्झ (मेगाहर्ट्ज) वा गिगाहर्ट्ज (GHz) मा मापन गरिएको छ। एक मेगाहेर्ट्ज प्रति सेकेन्ड १० लाख टिक्स बराबर हुन्छ, र एक गिगाहर्ट्ज बराबर एक सेकेन्ड टिक टिक प्रति सेकेन्ड।

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3 years ago

HELP ME !!!!!!<br> instruction given the quadrilateral ABCD inscribed in the circle, find x and y

grin007 [14]

The opposite angles in a cyclic quadrilateral are supplementary. i.e., the sum of the opposite angles is equal to 180˚

So using theorem we find that

∠A+∠C = 180

x + 108 = 180

x = 180 - 108

x = 72

∠B+∠D = 180

88 + y = 180

y = 180 - 88

y = 92

Answered by Gauthmath must click thanks and mark brainliest

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3 years ago

A ball is thrown from an initial height of 3 feet with an initial upward velocity of 21 ft/s. The ball height h (in feet) after

Alexxx [7]

Answer:

Step-by-step explanation:

The position equation is given as

$s(t)=-16t^2+21t+3$ and we are looking for the times when the position of the ball is 9 feet. That means that we simply plug in a 9 for s(t) and factor to solve for t:

$9=-16t^2+2t+3$ and

$0=-16t^2+21t-6$ and this is what we factor to find t:

t = .42 sec and then again on its way back down at t = .89 sec

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3 years ago

A thermometer is taken from an inside room to the outside, where the air temperature is 20° f. after 1 minute the thermometer re

fomenos

The change in the temperature is proportional to the difference in temperatures. Let's call the initial temperature of the room and the thermometer $E_0$ and the current temperature $E(t)$ where $t$ is the elapsed time in minutes. The outside temperature is 20 F.

Newton's Law of Cooling:

$E(t) - 20 = (E_0 - 20) e^{- k t}$

We're given $E(1)=70 \textrm{ and } E(5)=45$

$70-20 = (E_0 - 20) e^{-k}$

$45-20 =(E_0 - 20) e^{-5k}$

Dividing,

$\dfrac{50}{25}=e^{4k}$

$k = \dfrac 1 4 \ln 2$

Now,

$E(t) - 20 = (E_0 - 20) e^{- k t}$

$E_0 = 20 + (E(t) - 20)e^{k t}$

$E_0 = 20 + (E(1) - 20)e^{k}$

$E_0 = 20 + (70 - 20)e^{\ln 2/4}$

$E_0 = 20 + 50 \cdot 2^{1/4} \approx 79.46 \textrm{ degrees F}$

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4 years ago