All categories

2 years ago

Transform Linear. A2. Q1

Mathematics

1 answer:

belka [17]2 years ago

8 0

That's correct
answer is a
translated 3 units up

You might be interested in

PLEASE SOMEONE HELP ME

kow [346]

Answer:

1/5 is not greater than 2/7 because 1/5 = 7/35 and 2/7 = 10/35.

7 0

2 years ago

Read 2 more answers

At the theater, tickets for adults cost $4.75

shutvik [7]

X+y= 26 and 4.75x + 2.25y= 83.50

y=26-x( substitute this into second equation for y and solve for x)

4.75x + 2.25(26-x)= 83.50

4.75x + 58.5-2.25x= 83.50

x= 10

Now solve for y by substituting your answer for x

10+ y= 26

y=16

Therefore, 16 tickets were purchased for kids and 10 for adults.

7 0

2 years ago

How to solve this problem

Shkiper50 [21]

You have to ask the question with the right type of something.

6 0

2 years ago

Please Help!!! I need -14+6=-8 in a addition sentence

padilas [110]

Step-by-step explanation:

megan owes the bank 14 dollars, but she then deposits 6 dollars. she now owes the bank 8 dollars

6 0

2 years ago

hi, i dont undertand number 20 because i was absent in class today and i rerally need help, i will appraciate with the help, and

Mariulka [41]

Given:

The equation is,

$2\log _3x-\log _3(x-2)=2$

Explanation:

Simplify the equation by using logarthimic property.

$\begin{gathered} 2\log _3x-\log _3(x-2)=2 \\ \log _3x^2-\log _3(x-2)=2_{}\text{ \lbrack{}log(a)-log(b) = log(a/b)\rbrack} \\ \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \end{gathered}$

Simplify further.

$\begin{gathered} \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \\ \frac{x^2}{x-2}=3^2 \\ x^2=9(x-2) \\ x^2-9x+18=0 \end{gathered}$

Solve the quadratic equation for x.

$\begin{gathered} x^2-6x-3x+18=0 \\ x(x-6)-3(x-6)=0 \\ (x-6)(x-3)=0 \end{gathered}$

From the above equation (x - 6) = 0 or (x - 3) = 0.

For (x - 6) = 0,

$\begin{gathered} x-6=0 \\ x=6 \end{gathered}$

For (x - 3) = 0,

$\begin{gathered} x-3=0 \\ x=3 \end{gathered}$

The values of x from solving the equations are x = 3 and x = 6.

Substitute the values of x in the equation to check answers are valid or not.

For x = 3,

$\begin{gathered} 2\log _3(3^{})-\log _3(3-2)=2 \\ 2\log _33-\log _31=2 \\ 2\cdot1-0=2 \\ 2=2 \end{gathered}$

Equation satisfy for x = 3. So x = 3 is valid value of x.

For x = 6,

$\begin{gathered} 2\log _36-\log _3(6-2)=2 \\ 2\log _36-\log _34=2 \\ \log _3(6^2)-\log _34=2 \\ \log _3(\frac{36}{4})=2 \\ \log _39=2 \\ \log _3(3^2)=2 \\ 2\log _33=2 \\ 2=2 \end{gathered}$

Equation satifies for x = 6.

Thus values of x for equation are x = 3 and x = 6.

6 0

9 months ago