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2 years ago

Please help!! This is super important!!

Mathematics

1 answer:

MAVERICK [17]2 years ago

3 0

Answer:

tell me the choices so i can help you............

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Find the inverse of ????−7 −2

Masja [62]

Answer:

$A^{-1}=\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}$

Step-by-step explanation:

$A=\begin{bmatrix}-7 & -2\\ 4 & 1\end{bmatrix}$

$det\ A=-7\times 1-(-2)\times 4\\\Rightarrow det\ A=1$

$A^{-1}=\frac{1}{det\ A}\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}\\\Rightarrow A^{-1}=\frac{1}{1}\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}\\\Rightarrow A^{-1}=\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}$

$\therefore A^{-1}=\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}$

$A.A^{-1}=\begin{bmatrix}-7 & -2\\ 4 & 1\end{bmatrix}\times \begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}\\\Rightarrow A.A^{-1}=\begin{pmatrix}\left(-7\right)\cdot \:1+\left(-2\right)\left(-4\right)&\left(-7\right)\cdot \:2+\left(-2\right)\left(-7\right)\\ 4\cdot \:1+1\cdot \left(-4\right)&4\cdot \:2+1\cdot \left(-7\right)\end{pmatrix}\\\Rightarrow A.A^{-1}=\begin{pmatrix}1&0\\ 0&1\end{pmatrix}$

Hence, confirmed

6 0

3 years ago

The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E

nexus9112 [7]

Answer:

The projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

Step-by-step explanation:

Consider the provided projected rate.

$E'(t) = 12000(t + 9)^{\frac{-3}{2}}$

Integrate the above function.

$E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt$

$E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c$

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

$2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c$

$2000=-\frac{24000}{3}+c$

$2000=-8000+c$

$c=10,000$

Therefore, $E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

Now we need to find $\lim_{t \to \infty} E(t)$

$\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

$\lim_{t \to \infty} E(t)=10,000$

Hence, the projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

8 0

2 years ago

Lou has an account with $10,000 which pays 6% interest compounded annually. If to that account, Lou deposits $5,000 at the begin

katovenus [111]

Answer:

Option d. $22154 is the right answer.

Step-by-step explanation:

To solve this question we will use the formula $A=P(1+\frac{r}{n})^{nt}$

In this formula A = amount after time t

P = principal amount

r = rate of interest

n = number of times interest gets compounded in a year

t = time

Now Lou has principal amount on the starting of first year = 10000+5000 = $15000

So for one year $A=15000(1+\frac{\frac{6}{100}}{1})^{1\times1}$

$= 15000(1+.06)^{1}$

$= 15000(1.06)$ = $15900

After one year Lou added $5000 in this amount and we have to calculate the final amount he got

Now principal amount becomes $15900 + $ 5000 = $20900

Then putting the values again in the formula

$A=20900(1+\frac{\frac{6}{100}}{1})^{1\times1}$

$= 20900(1+.06)^{1}$

$= 20900(1.06)=22154$

So the final amount will be $22154.

3 0

2 years ago

If the Volume of a box is 250 feet3 and the height is 5 feet and the length is 5

Montano1993 [528]

Answer:

10 feet

Step-by-step explanation:

Use the volume formula, V = lwh, to solve for w

Plug in the values we know:

V = lwh

250 = (5)(w)(5)

250 = 25w

10 = w

So, the width is 10 feet

7 0

2 years ago

when a number is added to 1/5 of itself, the result is 24. the equation that models this problem is n 1/5 n = 24. what is the va

GuDViN [60]

Let n be unknown number, then 1/5 of this number is 1/5n. When you add these two numbers, you get

$n+\dfrac{1}{5} n.$

You know that this sum is equal to 24, so

$n+\dfrac{1}{5} n=24.$

1. Multiply this equation by 5:

$5n+n=120.$

2. Solve it:

$6n=120,\\n=120:6,\\n=20.$

Answer: n=20, correct choice is B.

4 0

2 years ago

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