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2 years ago

PLEASE I NEED HELP PLEASE

Mathematics

1 answer:

mr_godi [17]2 years ago

3 0

Answer:

none

Step-by-step explanation:

there's only ten cubes

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What is the greatest whole number that rounds to the number 277,300?

lorasvet [3.4K]

That whole number is 277,349.

8 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

PLEASE ANSWER CORRECTLY, DON'T JUST STEAL THE POINTS.

marin [14]

Since you're only asked the ordered pair of D'', it's much easier just to plot and reflect point D twice than to do that for all four points!

Remember that reflecting points is like putting a mirror at the line of reflection or flipping that point over at that line. The reflected point should be the same distance from the line of reflection as the original point.

1) Reflect D over the x-axis to get D'.
D is at (4,1). Draw a line that is perpendicular to the line of reflection and goes through D. D is as far from the line of reflection as D' should be on its other side (both are on that perpendicular line). Since D is 1 unit above the x-axis, that means D' is 1 unit below at (4, -1). See picture 1.

2) Reflect D' over y=x+1 to get D''.
D' is at (4, -1). Draw y=x+1 and the line perpendicular to it going through D''. D'' is the same distance from the line of reflection on the other side. See picture 2. D'' is at (-2, 5).

Answer: D'' is at (-2, 5)