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Thepotemich [5.8K]
3 years ago
10

In a certain quiz that consists of 10 questions, each question after the first is worth 4 points more than the preceding questio

n. If the 10 questions on the quiz are worth a total of 360 points, how many points is the third question worth?A. 18B. 24C. 26D. 32E. 44
Mathematics
1 answer:
weqwewe [10]3 years ago
8 0

Answer:

C. 26

Step-by-step explanation:

Let’s suppose that the first question is worth ‘’x’’ amount of points. Then the second question will be worth x+4, the third question x+8, the fourth question x+12, the fifth x+ 16, the sixth x+20, the seventh x+24, the eighth x+ 28, the ninth x+ 32, and the last one x + 36.

The sum of the points of all these questions is equal to 10x + 180. Since the quiz has a total of 360 points, then

10x + 180 = 360.

Solving for x we get:

X = 18.

This means that the first question is worth 18 points, the second one 22 points (18 +4), the third one 26 points(18+4+4), and so on.  

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I really need help with my test ill do anything plzz
lesantik [10]

Answer:

B

Step-by-step explanation:

All the other points seem to be present on the graph

4 0
3 years ago
Need answer thanks! only answer if you know please​!
Vaselesa [24]

Answer:

a) k \approx 0.0602\,\frac{1}{min} (k\approx  6.02\,\frac{\%}{min}), b) k \approx 0.0648\,\frac{1}{min} (k \approx 6.48\,\frac{\%}{min}), c) t_{1/2} \approx 60.274\,days, d) t_{1/2}\approx 4.305\,min, e) k \approx 0.3\,\frac{1}{min} (k\approx 30\,\frac{\%}{min}), f) k \approx 0.0120\,\frac{1}{min} (k \approx 1.20\,\frac{\%}{min}), g) k \approx 2.876\times 10^{-5}\,\frac{1}{yr} (k \approx 2.876\times 10^{-3}\,\frac{\%}{yr})

Step-by-step explanation:

All radioactive isotopes decay exponentially, the mass of the isotope as a function of time (m(t)), measured in grams, is defined by:

m(t) = m_{o}\cdot e^{-k\cdot t} (1)

Where:

m_{o} - Initial mass of the isotope, measured in grams.

k - Decay rate, measured in \frac{1}{min} or \frac{1}{day} or \frac{1}{yr}.

t - Time, measured in minutes, days or years.

We define the decay rate in terms of half-life by using the following expression:

k = \frac{\ln 2}{t_{1/2}} (2)

Where t_{1/2} is the half-life of the isotope, measured in minutes or years.

Now we proceed to determine the missing values:

a) Polonium-200 (t_{1/2} = 11.5\,min)

k = \frac{\ln 2}{11.5\,min}

k \approx 0.0602\,\frac{1}{min}

b) Lead-194 (t_{1/2} = 10.7\,min)

k = \frac{\ln 2}{10.7\,min}

k \approx 0.0648\,\frac{1}{min}

c) Iodine-125 (k = 0.0115\,\frac{1}{day})

t_{1/2} = \frac{\ln 2}{k}

t_{1/2} = \frac{\ln 2}{0.0115\,\frac{1}{day} }

t_{1/2} \approx 60.274\,days

d) Kryption-75 (k = 0.161\,\frac{1}{min})

t_{1/2} = \frac{\ln 2}{k}

t_{1/2} = \frac{\ln 2}{0.161\,\frac{1}{min} }

t_{1/2}\approx 4.305\,min

e) Strontium-79 (t_{1/2} = 2.3\,min)

k = \frac{\ln 2}{2.3\,min}

k \approx 0.3\,\frac{1}{min}

f) Uranium-229 (t_{1/2} = 58\,min)

k = \frac{\ln 2}{58\,min}

k \approx 0.0120\,\frac{1}{min}

g) Plutonium-239 (t_{1/2} = 24100\,yr)

k = \frac{\ln 2}{24100\,yr}

k \approx 2.876\times 10^{-5}\,\frac{1}{yr}

6 0
3 years ago
A rectangle piece of land is 400m*150m.Find the cost of the piece of land if one square meter of land costs ₹15,000.​
Andrew [12]

Answer:

900000000

Step-by-step explanation:

Area=400×500

=60000m²

Since

1m² = 15000

60000m² = x

x = 60000×15000

x = 900000000

5 0
3 years ago
Pleaseeee help me i have very little time
VashaNatasha [74]

Answer:

212 feet

Step-by-step explanation:

Please let me know if you want me to add an explanation as to why this is the answer. I can definitely do that, I just wouldn’t want to write it if you don’t want me to :)

7 0
3 years ago
I need help !!! Only 1 ATTEMPT please
astraxan [27]
Base and height so you would multiply 1/2 with 4 then divide by two and you get your answer
6 0
3 years ago
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