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3 years ago

Which situations described can be modeled by an exponential function?

Mathematics

2 answers:

nasty-shy [4]3 years ago

7 0

Answer:

I did this exact same question on USATestPrep and here's what I answered that I got correct:

B, C and this one isn't an answer choice but the additional answer was: "1,000 grams of radioactive material with a half-life of 6 hours."

Hope this helps! I'm sorry if it doesn't help future people who sees this :/

seropon [69]3 years ago

7 0

Answer:

B, C, or the additional answer that was said above. if i got this wrong would someone let me know in the comments. thank you!

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Reduce the following fraction to lowest terms:

Svetradugi [14.3K]

It would end up being 4/5 if you brought it to the lowest tern.

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3 years ago

Where do the equations y=-4x+8 and y=3x+1 meet

Dafna11 [192]

Answer:

l think they meet when collecting like terms.maybe like this

y=4x8

y=3x+1

=(4×+3x)+1+8

7 0

3 years ago

PLEASE ANSWER ASAP! SUPER IMPORTANT! GIVING POINTS!

katen-ka-za [31]

-13 +16 = 3=odd+even = odd

-7 + 16 = 9 =odd+even = odd

4 0

3 years ago

Prove that 2n^3 + 3n^2 + n is divisible by 6 for every integer n > 1 by mathematical induction. (7 marks)

notka56 [123]

Answer:

The answer is True

Step-by-step explanation:

A mathematical induction consists in only 2 steps:

First step: Show the proposition is true for the first one valid integer number.

Second step: Show that if any one is true then the next one is true

Finally, if first step and second step are true, then the complete proposition is true.

So, given $2*n^3+3*n^2+n$

First step: using and replacing n=2 (the first valid integer number >1)

$2*(2)^3 +3*(2)^2+2=30$

$\frac{30}{6} =5$

As the result is an integer number, so the first step is true.

Second step: using any next number, $n+1$ , let it replace

$2*(n+1)^3+3*(n+1)^2+(n+1)\\2*(n^3+3*n^2+3*n+1)+3*(n^2+2*n+1)^2+(n+1)\\2*n^3+6*n^2+6*n+2+3*n^2+6*n+3+n+1\\(2*n^3+3*n^2+n)+(6*n^2+12*n+6)$

As the First step is true, we know that

$2*n^3+3*n^2+n$ $=6*k$ ,

So let it replace in the previous expression

$6*k+6*(n^2+2*n+1)\\6*[k+(n^2+2*n+1)]$

Finally

$\frac{6*[k+(n^2+2*n+1)]}{6} =k+(n^2+2*n+1)$

where the last expression is an integer number

So the second step is true, and the complete proposition is True

5 0

3 years ago

(2^3)^7 (2^-9)^2 how do i put that in exponential form

jok3333 [9.3K]

Answer:

2³ = 8

Step-by-step explanation:

According to the laws of exponents, when raising a power to another power, you need to multiply the exponents. After you multiply each set of exponents by the additional power, then it become a multiplication problem where the base is the same. When multiplying exponents with the same base, you keep the base and add the exponents:

$(2^{3})^{7}*(2^{-9})^{2}=2^{21}*2^{-18}=2^{21+-18}=2^{3}$

7 0

3 years ago