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emmasim [6.3K]
3 years ago
8

Solve the system.

Mathematics
1 answer:
gtnhenbr [62]3 years ago
8 0
X = y + 11

2x + y = -2
2(y + 11) + y = -2
2y + 22 + y = -2
3y + 22 = -2
3y = -24
y = -8

x = y + 11
x = -8 + 11
x = 3

Solution set {3, -8} (C)
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Felix's parents opened a savings
Crazy boy [7]

The money in the Felix's account will be $6798 when he is 21.

<u>Step-by-step explanation:</u>

It is given that,

  • The amount deposited is $2000.
  • The account earns 6%  compound interest.
  • It is compounded  annually for 21 years.

<u>To find the money in Felix's account after 21 years :</u>

The formula used here is,

⇒ A = P(1+r/n)^{nt}

where A is the amount after 21 years.

  • P is the initial amount deposited ⇒ P = 2000
  • r is the rate ⇒ r = 0.06
  • n is the number of times interest is compounded per year⇒ n = 1
  • t is the time period ⇒ t = 21

⇒ 2000(1+0.06/1)^{21\times 1}

⇒ 2000(1.06)^{21}

⇒ 2000\times3.399

⇒ 6798

Therefore, The money in the Felix's account will be $6798 when he is 21.

4 0
4 years ago
Find the value of x if y = -14, m = -2, and b = -2 in the formula y = mx + b.
posledela

Answer:

Step-by-step explanation:

y = -14, m = -2, b = -2

-14 = -2(x) + -2

-14 = -2x + -2

-12 = -2x

Divide by -2.

x = 6

8 0
3 years ago
A model of a submarine has a length of 6.6 meters. Use the scale 1:35 to find the length
kirill115 [55]
The scale 1:35 indicates that for every 1 meter on the scale model, there are really 35 meters in real life.

So if the scale model is 6.6 meters, you would do 35 * 6.6 to find the length in real life, which is 231 meters.

The answer is 231 meters. 
7 0
3 years ago
Help please! i don’t understand how to do this!
atroni [7]

Answer:

x=c+b/a

Step-by-step explanation:

ax-b=c

    +b +b

ax=c+b

__ ___

a=c+b/a

4 0
3 years ago
Read 2 more answers
Exercise 3.7.4: let a = 2 1 0 0 2 0 0 0 2 .
swat32

With

\mathbf A=\begin{bmatrix}2&1&0\\0&2&0\\0&0&2\end{bmatrix}

we have

\det(\mathbf A-\lambda\mathbf I)=\begin{vmatrix}2-\lambda&1&0\\0&2-\lambda&0\\0&0&2-\lambda\end{vmatrix}=(2-\lambda)^3

so \mathbf A has one eigenvalue, \lambda=2, with multiplicity 3.

In order for \mathbf A to not be defective, we need the dimension of the eigenspace to match the multiplicity of the repeated eigenvalue 2. But \mathbf A-2\mathbf I has nullspace of dimension 2, since

\begin{bmatrix}0&1&0\\0&0&0\\0&0&0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\mathbf 0\implies x=0\text{ or }y=0

That is, we can only obtain 2 eigenvectors,

\begin{bmatrix}1\\0\\0\end{bmatrix}\text{ and }\begin{bmatrix}0\\0\\1\end{bmatrix}

and there is no other. We needed 3 in order to complete the basis of eigenvectors.

3 0
3 years ago
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