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3 years ago

A rectangle has an area of 120cm2. It's length and width are whole numbers.

1 answer:

7 0

The two numbers can be any factors of 120. This includes: (1,120) (2,60)(3,40)(4,30)(5,24)(6,20)(8,15)(10,12) and the reverse of this. The smallest perimeter would be the two numbers that are closer together, which is 10 and 12.

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The answer to this equation

12345 [234]

Answer:

x=3

Step-by-step explanation:

We are given the equation

$log5x=log(2x+9)$

This can be rewritten as

$log5x-log(2x+9)=0$

Now using the quotient rule for logarithms we can combine these two

$log(\frac{5x}{2x+9} )=0$

Next we can remove the log by using an inverse operation

$10^{log(\frac{5x}{2x+9} )} =10^0\\\\\frac{5x}{2x+9}=1$

Now we can solve for x

$\frac{5x}{2x+9}=1\\\\5x=2x+9\\\\3x=9\\\\x=3$

4 0

3 years ago

Click on the numbers to enter the answers in the boxes.

djyliett [7]

Answer:

957 - 249 = 708

The number that goes above the 5 is 4

Number above the 7 is 17

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Directions- Solve the following problem and write your answer in scientific notation. The mass of the moon is about 7.3X10^22 kg

Masteriza [31]

Answer:

Step-by-step explanation:

7.3×10^22 = 73,000,000,000,000,000,000,000

73,000,000,000,000,000,000,000 × 26,000,000 = 1,898,000,000,000,000,000,000,000,000,000 or 1.898 × 10^30

3 0

3 years ago

Read 2 more answers

Rewrite 3/9 as a mixed number

Daniel [21]

3/9 is already a fraction, you do not need to change it. 3/9 is not more than 9/9

4 0

3 years ago

Read 2 more answers

For the function given state the period f(t) =6sin(3t-pi/6)-1

lapo4ka [179]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}sin({{ B}}x+{{ C}})+{{ D}}
\\\\
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}\\\\
f(x)=&{{ A}}tan({{ B}}x+{{ C}})+{{ D}}

\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\bf \begin{array}{llll}


\bullet \textit{vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{function period or frequency}\\
\qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\
\qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)
\end{array}$

now, with that template in mind, let's take a peek at yours

$\bf \begin{array}{lllcclll}
f(t)=&6sin(&3t&-\frac{\pi }{6})&-1\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}\\\\
-----------------------------\\\\
period\qquad \cfrac{2\pi }{B}\iff\cfrac{2\pi }{3}$

8 0

3 years ago