1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
tia_tia [17]
3 years ago
8

Really need the answer to this.

Mathematics
1 answer:
vredina [299]3 years ago
7 0

Answer:


Step-by-step explanation:


You might be interested in
How many inches are in one yard? (Show your work
Aleonysh [2.5K]
36 inches are in a yard
8 0
2 years ago
Read 2 more answers
Use the given inverse to solve the system of equations. left brace Start 3 By 3 Matrix 1st Row 1st Column x minus y plus z 2nd C
natka813 [3]

The interpretation of the given question is as follows:

Use the given inverse to solve the system of equations

x- y - z = -6  \\ \\ 2y + z = -6 \\ \\  3x -8 y = - \dfrac{1}{2}

The inverse of  \left[\begin{array}{ccc}1&-1&1\\0&2&1\\3&-8&0\end{array}\right]   is \left[\begin{array}{ccc}-8&8&3\\-3&3&1\\6&-5&-2\end{array}\right]

x =

y =

z =

Answer:

x = - 1.5

y = - 0.5

z =  - 5

Step-by-step explanation:

Using the correlation  of inverse of matrix AX = B to solve the question above;

AX = B

⇒ A⁻¹(AX)  = A⁻¹ B

X =  A⁻¹ B

So ;

X         =        A⁻¹                             B

\left[\begin{array}{c}x\\y\\z\end{array}\right] =     \left[\begin{array}{ccc}-8&8&3\\-3&3&1\\6&-5&-2\end{array}\right] =   \left[\begin{array}{ccc}-6\\ -6\\- \dfrac{1}{2}\end{array}\right]

   

 \left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}(-8*-6)+(8*-6)+(3*-\dfrac{1}{2})\\(-3*-6)+(3*-6)+(1*-\dfrac{1}{2})\\(6*-6)+(5*-6)+(-2* - \dfrac{1}{2})\end{array}\right]

\left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}(48)+(-48)+(\dfrac{-3}{2})\\(18)+(-18)+(\dfrac{-1}{2})\\(-36)+(30)+(1)\end{array}\right]

\left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}(\dfrac{-3}{2})\\(\dfrac{-1}{2})\\(-5)\end{array}\right]

\left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}-1.5\\-0.5\\ -5\end{array}\right]

5 0
3 years ago
A bag contains 2 red marbles, 7 blue marbles and 5 green marbles. If two marbles are drawn out of the bag, what is the exact pro
vivado [14]

Answer:

0.25 or 25%.

Step-by-step explanation:

2 Red + 3 Blue + 7 Green = 12 Total Marbles.

You have 3 blue marbles out of the total marbles, so 3/12, then simplify it. 3/12 = 1/4. This gives you 0.25, or 25%.  

The probability of drawing a blue marble is 0.25 or 25%

4 0
3 years ago
(UFRGS) Se 10x = 20y , atribuindo 0,3 para log2, então o valor de é
sveticcg [70]

Answer:

\frac{x}{y}=1.3

Step-by-step explanation:

<u>Informação recuperada:</u>

(UFRGS) Se 10x = 20y , atribuindo 0,3 para log 2 , então o valor de x/y é

1) Primeiro, passando o 10 para o 2º membro como base do logaritmo:

10x=20y \Rightarrow x=log_{10}20y

2) Aplicando a propriedade do produto de logaritmo:

log(AB)= logA+logB

x=log_{10}20y \Rightarrow x=log(2+10)y\Rightarrow x=(0.3+1)y \Rightarrow x=1.3y

3)  Como quero o quociente divido ambos os lados por y

\frac{x}{y}=\frac{1.3y}{y}\Rightarrow \frac{x}{y}=1.3

4 0
3 years ago
Find the domain and range
tresset_1 [31]
Domain: all real numbers
Range: y>-3
4 0
2 years ago
Other questions:
  • Is this a linear function?
    11·2 answers
  • Marshall is selling raffle tickets to raise money.he want to raise $1000 he already has $200 in donations,and each ticket will r
    5·2 answers
  • Can you guys help me?​
    13·2 answers
  • What times what equals 162
    13·1 answer
  • A baseball team wins 70% of its games.
    6·2 answers
  • PLEASE HELP!! thank you!!!
    15·1 answer
  • What are the odds of rolling a multiple of 5 using a 20 sided die?
    15·1 answer
  • In a circle of any size, what ratio does
    15·2 answers
  • Help pls and ty! What is the area of this trapezoid?
    5·2 answers
  • Find the missing angles.
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!