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Yanka [14]
3 years ago
11

Which statement correctly maps AB to A′B′

Mathematics
1 answer:
VARVARA [1.3K]3 years ago
8 0
Answer: ( x + 8, y -7)
x value is moved 8 units to the left. So add 8 to x value.
Y value is moved 7 units down. So subtract 7 from y value
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A kangaroo hops 2 kilometers in 3 minutes. At this rate, how long does it take the kangaroo to travel 5 kilometers?
Mandarinka [93]

Answer: 7.5 minutes (7 min 30 sec)

Step-by-step explanation: If it takes it 3 minutes to travel 2 km, it would take it 1.5 minutes to travel 1 km. So, 1.5 * 5 = 7.5, meaning it would take it 7.5 min to travel 5 km.

7 0
3 years ago
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Examine the expanded form.<br><br> a ∙ a ∙ a ∙ a ∙ a ∙ a ∙ a
Juliette [100K]

Answer:

a^7

Step-by-step explanation:

a to the 7th power

6 0
4 years ago
27 ÷ (−9) =<br><br> give step by step explaination<br>​
maria [59]

Answer:

-3

Step-by-step explanation:

27 divided by 9 is 3

the signs (+) / (-) = -

6 0
3 years ago
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There are 4 fewer cats than dogs.<br>There are 2 cats. How many dogs<br>are there?​
allsm [11]

Answer:

there are 6 dogs

Step-by-step explanation:

x = dogs

x= 2 + 4

x = 6

8 0
3 years ago
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Carbon-14 is a radioactive isotope that has a half-life of 5,730 years. Approximately how many years will it take for carbon-14
PilotLPTM [1.2K]

Answer:

<em>Carbon-14 will take 19,035 years to decay to 10 percent.</em>

Step-by-step explanation:

<u>Exponential Decay Function</u>

A radioactive half-life refers to the amount of time it takes for half of the original isotope to decay.

An exponential decay can be described by the following formula:

N(t)=N_{0}e^{-\lambda t}

Where:

No  = The quantity of the substance that will decay.

N(t) = The quantity that still remains and has not yet decayed after a time t

\lambda     = The decay constant.

One important parameter related to radioactive decay is the half-life:

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

If we know the value of the half-life, we can calculate the decay constant:

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

Carbon-14 has a half-life of 5,730 years, thus:

\displaystyle \lambda=\frac {\ln(2)}{ 5,730}

\lambda=0.00012097

The equation of the remaining quantity of Carbon-14 is:

N(t)=N_{0}e^{-0.00012097\cdot t}

We need to calculate the time required for the original amout to reach 10%, thus N(t)=0.10No

0.10N_o=N_{0}e^{-0.00012097\cdot t}

Simplifying:

0.10=e^{-0.00012097\cdot t}

Taking logarithms:

ln 0.10=-0.00012097\cdot t

Solving for t:

\displaystyle t=\frac{log 0.10}{-0.00012097}

t\approx 19,035\ years

Carbon-14 will take 19,035 years to decay to 10 percent.

6 0
3 years ago
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