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3 years ago

Six times the difference of a number and 9 is 54. Find the number.

Mathematics

2 answers:

Goryan [66]3 years ago

7 0

Answer: 6

Step-by-step explanation:

9 x 6 =54

54/9 =6

Goshia [24]3 years ago

4 0

Answer:

x=18

Step-by-step explanation:

6(x-9)=54

Divide by 6 on each side

x-9=9

Add 6 on both sides

x=18

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What is the interquartile range of this data? 6 8 9 20

givi [52]

Answer:

Step-by-step explanation:

The interquartile range is the difference between the values of the upper quartile and the lower quartile.

The upper quartile is the value at the right side of the box, that is 36

The lower quartile is the value at the left side of the box, that is 16

Interquartile range = 36 - 16 = 20

6 0

3 years ago

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Solve: 2 3 x - 4 = -2

andrew11 [14]

Hi there!

23x - 4 = -2
23x = 2
x = 2/23

Hope this helps!

6 0

3 years ago

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10 POINTS PLS ANSWER

Marina CMI [18]

Answer:

x=15

Step-by-step explanation:

x+165degrees=180degrees

x=180-165

x=15

6 0

3 years ago

Simplify the expression completely. (5√50-3√200+3√18)

Kaylis [27]

Answer:

$4\sqrt{2}$

Step-by-step explanation:

$5\sqrt{50} - 3\sqrt{200} + 3\sqrt{18}$ =

= $25\sqrt{2} - 30\sqrt{2} + 9\sqrt{2}$ = $4\sqrt{2}$

8 0

3 years ago

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I need help with #11

Troyanec [42]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\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{ period of }\frac{2\pi }{{{ B}}}
\end{array}$

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

$\bf \begin{array}{lllcclll}
y=&2(&1x&-2)^2&-4\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}\\\\
-----------------------------\\\\
A\cdot B=2\impliedby \textit{shrunk by a factor of 2, of half-size}\\\\
\cfrac{C}{B}= \cfrac{-2}{1}\implies -2\impliedby \textit{horizontal right shift of 2 units}\\\\
D=-4\impliedby \textit{vertical down shift of 4 units}$

so, the graph of y=2(x-2)²-4, is really the same graph of y=x², BUT, narrower, and moved about horizontally and vertically

8 0

3 years ago

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