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3 years ago

What is 2+2736=344×2484848÷484848=

Mathematics

1 answer:

hammer [34]3 years ago

8 0

2+2736=2738*2484848/484848=14032.26129

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A=1/2f(r+z), solve for r

marin [14]

Answer: $\frac{2A}{f} -2=f$

Step-by-step explanation:

A=1/2(r+2)

2A=f(r+2)

2A/f=r+2

2A/f-2=r

7 0

2 years ago

Evaluate A/B for a = 1/2 and b = -3/7

bonufazy [111]

Answer:

-7/6

Step-by-step explanation:

If a = 1/2 and b = -3/7, then your given:

1/2 divided by -3/7=

-7/2*3=

-7/6

Sorry if its a bit unclears

4 0

3 years ago

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7) Original price: ?<br> Discount: 25%<br> Sale price:<br> $28

Studentka2010 [4]

Answer:

Step-by-step explanation:

let be x the price before the discount

original price= 28/(1-0.25)=28/0.75=37.33

original price=37.33

3 0

3 years ago

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Please graph y= 1/2x–3<br> Please i need help with this <3

erik [133]

Answer:

coordinates are 6, -3

Step-by-step explanation:

When it's plotted out onto a graph, the answer comes down to 6, -3 (also remember, it's X then Y)

6 0

3 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

1 year ago