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3 years ago

What is the coefficients in the following expression

Mathematics

2 answers:

Nikolay [14]3 years ago

6 0

Answer:

Step-by-step explanation:

MAXImum [283]3 years ago

4 0

Answer:

Step-by-step explanation:

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HELP PLEASE

frozen [14]

Answer:

it would take seven and a half hours

Step-by-step explanation:

450 ÷60 =7.5

8 0

3 years ago

Find the length of the arc and express your answer as a fraction times pie

Elina [12.6K]

Solution:

Given a circle of center, A with radius, r (AB) = 6 units

Where, the area, A, of the shaded sector, ABC, is 9π

To find the length of the arc, firstly we will find the measure of the angle subtended by the sector.

To find the area, A, of a sector, the formula is

$\begin{gathered} A=\frac{\theta}{360\degree}\times\pi r^2 \\ Where\text{ r}=AB=6\text{ units} \\ A=9\pi\text{ square units} \end{gathered}$

Substitute the values of the variables into the formula above to find the angle, θ, subtended by the sector.

$\begin{gathered} 9\pi=\frac{\theta}{360\degree}\times\pi\times6^2 \\ Crossmultiply \\ 9\pi\times360=36\pi\times\theta \\ 3240\pi=36\pi\theta \\ Divide\text{ both sides by 36}\pi \\ \frac{3240\pi}{36\pi}=\frac{36\pi\theta}{36\pi} \\ 90\degree=\theta \\ \theta=90\degree \end{gathered}$

To find the length of the arc, s, the formula is

$\begin{gathered} s=\frac{\theta}{360\degree}\times2\pi r \\ Where \\ \theta=90\degree \\ r=6\text{ units} \end{gathered}$

Substitute the variables into the formula to find the length of an arc, s above

$\begin{gathered} s=\frac{\theta}{360}\times2\pi r \\ s=\frac{90\degree}{360\degree}\times2\times\pi\times6 \\ s=\frac{12\pi}{4}=3\pi\text{ units} \\ s=3\pi\text{ units} \end{gathered}$

Hence, the length of the arc, s, is 3π units.

4 0

1 year ago

(1/2)^2 - 6 (2 - 2/3)<br><br> Note: 1/2 and 2/3 is a fraction

shusha [124]

$\bf \left( \cfrac{1}{2} \right)^2-6\left(2-\cfrac{2}{3} \right)\impliedby recall~~\mathbb{PEMDAS}
\\\\\\
\cfrac{1^2}{2^2}-6\left( \cfrac{6-2}{3} \right)\implies \cfrac{1}{4}-6\left( \cfrac{4}{3}\right)\implies \cfrac{1}{4}-\cfrac{6\cdot 4}{3}\implies \cfrac{1}{4}-\cfrac{24}{3}
\\\\\\
\cfrac{1}{4}-8\impliedby LCD~~4\implies \cfrac{1-32}{4}\implies \cfrac{-31}{4}\implies -7\frac{3}{4}$

5 0

3 years ago

How does the graph of f(x)=3lx+2l+4 relate to its parent function?

Sergeeva-Olga [200]

$\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 above in mind, let's see this one

$\bf parent\implies f(x)=|x|
\\\\\\
\begin{array}{lllcclll}
f(x)=&3|&1x&+2|&+4\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}$

A=3, B=1, shrunk by AB or 3 units, about 1/3
C=2, horizontal shift by C/B or 2/1 or just 2, to the left
D=4, vertical shift upwards of 4 units

check the picture below

7 0

3 years ago

Select a counter-example that makes the conclusion false.

Softa [21]

A counter-example could be

There are more than 3 marbles in the bag, and the other ones are red.

Hope this helps!

4 0

3 years ago