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astra-53 [7]
2 years ago
9

In a triangle ABC,AB=9 and BC =12 which of the following Cannot be the length of AC.

Mathematics
1 answer:
galben [10]2 years ago
5 0

Step-by-step explanation:

mark it as the brainliest

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Find a logarithmic equation that relates y and x
Svet_ta [14]
Logarithms are only able to make an equation linear if it is an exponential function.  For example, y= e^{x} can be made linear by taking the natural logarithm of each side, causing it to become ln(y)=ln( e^{x}). After some simplifying, you are left with ln(y)=x. You are then able to plot ln(y) vs. x to get a linear fit. 
4 0
1 year ago
PLZZZZZZ HELP WILL GIVE BRAINLEST
jonny [76]

−7⋅(−3)⋅(−9)⋅(−3)

21 * -9 * -3

-189 * -3

567

Choice D

4 0
2 years ago
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The sanitation department calculated that last year each city resident produced approximately 1.643 × 103 pounds of garbage. The
bija089 [108]

Given that, according to sanitation departments report; the samount of garbage produced by 1 resident = 1.643 \times 10^3pound.

Given that population of the city = 2.61 \times 10^5


Now we have to find how much garbage did the city sanitation department collect last year.

To find that we just need to multiply the garbage produced by 1 resident with the population of the city.


Total garbage collected = (2.61 \times 10^5 ) * (1.643 \times 10^3 ) pounds

Total garbage collected = (2.61*1.643) \times 10^{5+3} pounds

Total garbage collected = 4.28823 \times 10^8 pounds

Hence final answer is 4.28823 \times 10^8 pounds.

4 0
2 years ago
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Each locker is shaped like a rectangular prism. Which has more storage space? explain/
seropon [69]
Locker 1 has more storage space because it has a greater volume.
The volume of locker 1 is 8,640 (48x15x12)
The volume of locker 2 is 7,200 (60x10x12)

I hope this helps.
4 0
2 years ago
Eli invested $ 330 $330 in an account in the year 1999, and the value has been growing exponentially at a constant rate. The val
Cloud [144]

Answer:

The value of the account in the year 2009 will be $682.

Step-by-step explanation:

The acount's balance, in t years after 1999, can be modeled by the following equation.

A(t) = Pe^{rt}

In which A(t) is the amount after t years, P is the initial money deposited, and r is the rate of interest.

$330 in an account in the year 1999

This means that P = 330

$590 in the year 2007

2007 is 8 years after 1999, so P(8) = 590.

We use this to find r.

A(t) = Pe^{rt}

590 = 330e^{8r}

e^{8r} = \frac{590}{330}

e^{8r} = 1.79

Applying ln to both sides:

\ln{e^{8r}} = \ln{1.79}

8r = \ln{1.79}

r = \frac{\ln{1.79}}{8}

r = 0.0726

Determine the value of the account, to the nearest dollar, in the year 2009.

2009 is 10 years after 1999, so this is A(10).

A(t) = 330e^{0.0726t}

A(10) = 330e^{0.0726*10} = 682

The value of the account in the year 2009 will be $682.

4 0
2 years ago
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