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3 years ago

Please help ASAP!!!!!!!!!!!

Mathematics

1 answer:

Effectus [21]3 years ago

4 0

Answer:

x = 3.76 ft

Step-by-step explanation:

To find the missing side, "x", solve using proportions.

In similar polygons, the 'new' polygon was created by multiplying every side of the 'old' polygon by the same number. Every side was multiply by the <u>scale factor</u>, "k", which tells you how much a polygon grew (k>1) or shrunk (0<k<1).

Therefore, if you divided the pairs of corresponding sides, they would be equal.

From the diagram, you can tell that the following sides correspond:

CB ~ GF

AB ~ EF

<u>Use the proportion</u>

$\frac{EF}{AB} = \frac{GF}{CB}$

$\frac{x}{6ft} = \frac{3.3ft}{5ft}$ Substitute values from diagram

$6ft*\frac{x}{6ft} = 6ft*\frac{3.3ft}{5ft}$ Multiply both sides by 6 ft

$x = \frac{6ft*3.3ft}{5ft}$ "x" is isolated. Multiply fraction by combining into numerator

$x = \frac{19.8ft}{5ft}$ Solved numerator. Divide for decimal answer.

$x = 3.76ft$

The value of 'x' is 3.76 ft.

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Answer:

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Step-by-step explanation:

So I set up my problem like this(This is a way to figure a certain amount of a percent or vice versa):

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3 years ago

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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.

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Simplify further.

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

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

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.

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

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.

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goldenfox [79]

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