All categories

2 years ago

Evaluate each expression. Determine if the final simplified form of the expression is positive or negative -42 (-4)2 42

Mathematics

1 answer:

Sati [7]2 years ago

8 0

Answer:

Given the expression:

$-4^2$

we know:

$4^2 = 4 \times 4 = 16$

then;

$-4^2 = -16$

Therefore, the value of expression $-4^2$ is -16 i.e negative.

$(-4)^2$

we know:

$4^2 = 4 \times 4 = 16$

$(-1)^2 = 1$

then;

$(-4)^2 = (-1)^2 \cdot (4)^2= 16$

Therefore, the expression $(-4)^2$ is 16 i.e Positive.

$4^2$

we know:

$4^2 = 4 \times 4 = 16$

then;

$4^2 = 16$

Therefore, the expression $4^2$ is 16 i.e Positive.

You might be interested in

What is the slop of a line perpendicular to the line whose equation is 4x+6y=108. Reduce your answer

3241004551 [841]

I honestly don’t know so don’t pay attention to me pls I’m so sorry

6 0

3 years ago

Compute the sum:

Nady [450]

You could use perturbation method to calculate this sum. Let's start from:

$S_n=\sum\limits_{k=0}^nk!\\\\\\\(1)\qquad\boxed{S_{n+1}=S_n+(n+1)!}$

On the other hand, we have:

$S_{n+1}=\sum\limits_{k=0}^{n+1}k!=0!+\sum\limits_{k=1}^{n+1}k!=1+\sum\limits_{k=1}^{n+1}k!=1+\sum\limits_{k=0}^{n}(k+1)!=\\\\\\=1+\sum\limits_{k=0}^{n}k!(k+1)=1+\sum\limits_{k=0}^{n}(k\cdot k!+k!)=1+\sum\limits_{k=0}^{n}k\cdot k!+\sum\limits_{k=0}^{n}k!\\\\\\(2)\qquad \boxed{S_{n+1}=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n}$

So from (1) and (2) we have:

$\begin{cases}S_{n+1}=S_n+(n+1)!\\\\S_{n+1}=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n\end{cases}\\\\\\
S_n+(n+1)!=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n\\\\\\
(\star)\qquad\boxed{\sum\limits_{k=0}^{n}k\cdot k!=(n+1)!-1}$

Now, let's try to calculate sum $\sum\limits_{k=0}^{n}k\cdot k!$ , but this time we use perturbation method.

$S_n=\sum\limits_{k=0}^nk\cdot k!\\\\\\
\boxed{S_{n+1}=S_n+(n+1)(n+1)!}\\\\\\
$

but:

$S_{n+1}=\sum\limits_{k=0}^{n+1}k\cdot k!=0\cdot0!+\sum\limits_{k=1}^{n+1}k\cdot k!=0+\sum\limits_{k=0}^{n}(k+1)(k+1)!=\\\\\\=
\sum\limits_{k=0}^{n}(k+1)(k+1)k!=\sum\limits_{k=0}^{n}(k^2+2k+1)k!=\\\\\\=
\sum\limits_{k=0}^{n}\left[(k^2+1)k!+2k\cdot k!\right]=\sum\limits_{k=0}^{n}(k^2+1)k!+\sum\limits_{k=0}^n2k\cdot k!=\\\\\\=\sum\limits_{k=0}^{n}(k^2+1)k!+2\sum\limits_{k=0}^nk\cdot k!=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\\\\\\
\boxed{S_{n+1}=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n}$

When we join both equation there will be:

$\begin{cases}S_{n+1}=S_n+(n+1)(n+1)!\\\\S_{n+1}=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\end{cases}\\\\\\
S_n+(n+1)(n+1)!=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\\\\\\\\
\sum\limits_{k=0}^{n}(k^2+1)k!=S_n-2S_n+(n+1)(n+1)!=(n+1)(n+1)!-S_n=\\\\\\=
(n+1)(n+1)!-\sum\limits_{k=0}^nk\cdot k!\stackrel{(\star)}{=}(n+1)(n+1)!-[(n+1)!-1]=\\\\\\=(n+1)(n+1)!-(n+1)!+1=(n+1)!\cdot[n+1-1]+1=\\\\\\=
n(n+1)!+1$

So the answer is:

$\boxed{\sum\limits_{k=0}^{n}(1+k^2)k!=n(n+1)!+1}$

Sorry for my bad english, but i hope it won't be a big problem :)

8 0

3 years ago

In 2016, the population of the United States was approximately 3.24 x 10^8. It is expected that the population will be about 3.9

aleksandrvk [35]

Answer:

7.4 x 10^7

Step-by-step explanation:

3.98 x 10^8 - 3.24 x 10^8 = 0.74 x 10^8 (not scientific notation, because it isn't full number)

0.74 x 10^8 ---> 7.4 x 10^7

- got seven by counting the spaces it moved over (right decrease, while left increases)

5 0

3 years ago

Zoologists are studying two newly discovered species of insects in a previously unexplored section of rain forest. They estimate

salantis [7]

The population Pa of insect A after t years is given by the equation
Pa = 1.3(1-0.038)^t
while the population Pb of insect B after t years is
Pb = 2.1(1-0.046)^t

We equate the above expressions to find the number of years t it will take the two populations to be equal:
Pa = Pb
1.3(1-0.038)^t = 2.1(1-0.046)^t
1.3(0.962)^t = 2.1(0.954)^t
These are the equations that can be used to determine how long it will be before the populations of the two species are equal.

We can now solve for t:
(0.962)^t / (0.954)^t = 2.1/1.3
(0.962/0.954)^t = 2.1/1.3
After taking the log of both sides of our equation, number of years t is
t = log (2.1/1.3) / log (0.962/0.954)
t = 57 years
Therefore, it will take 57 years for the population of insect A to equal the population of insect B.

8 0

2 years ago

Read 2 more answers

Martin deposits $700 into a savings account that gains interest at a rate of 5.5% annually. If he waits 7 years to withdraw the

Travka [436]

｡☆✼★ ━━━━━━━━━━━━━━ ☾

Use the multiplier 1.055

final = original x multiplier^n

where 'n' is the number of years

Sub the values in

final = 700 x 1.055^7

final = 1018.27

Subtract the values:

1018.27 - 700 = 318.27

Thus, he earns more than $275

He earns $43.27 more

Have A Nice Day ❤

Stay Brainly! ヅ

- Ally ✧

｡☆✼★ ━━━━━━━━━━━━━━ ☾

4 0

2 years ago