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3 years ago

Without multiplying order the following products from least to greatest

Mathematics

2 answers:

AVprozaik [17]3 years ago

7 0

Answer:

It would be:

4 5/9 x 1/2

4 5/9 x 10/10

4 5/9 x 6

Step-by-step explanation:

Leno4ka [110]3 years ago

4 0

2, 3, 1

Hope this helps :)

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find three consecutive odd integers such that the sum of the smallest number and middle number is 27 less than 3 times the large

katrin2010 [14]

A generic odd number can be written as

$2k+1,\quad k \in \mathbb{Z}$

Since there is an odd number every two numbers, three consecutive odd numbers will be

$2k+1,\quad 2k+3,\quad 2k+5$

Now let's make up the equations: the sum of the first two is

$(2k+1)+(2k+3)$

And 27 less than 3 times the largest is

$3(2k+5)-27$

These two must be the same, so we have

$(2k+1)+(2k+3)=3(2k+5)-27 \iff 4k+4 = 6k+30-27 \iff 4k+4=6k+3$

Subtracting 4k and 3 from both sides gives

$1=2k \iff k=\dfrac{1}{2}$

Which means that the problem has no solution.

To confirm this hypothesis, we can observe that, on the left hand side, we have the sum of two odd numbers, which is even

On the right hand side, we have an odd number, multiplied by 3 (still odd), take away 27 (still odd).

So, the left hand side is even, and the right hand side is odd. They can't be the same number.

7 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

10 months ago

WILL DO BRAINLIEST!!!! Suppose the store wants to earn a daily profit of $150 from the sale of soccer balls. To earn this profit

lara [203]

Think about it this way, what is 5x3??? 15 right. Ok then what is 5x30???? 150. So the store should sell each soccer ball for $5 each:)

6 0

3 years ago

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slava [35]