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Softa [21]
3 years ago
15

If I can paint 96 sq ft in 15 minutes how long does it take to paint 1 sq ft

Mathematics
1 answer:
White raven [17]3 years ago
8 0

It takes 0.15625 minutes to paint 1 square feet

<em><u>Solution:</u></em>

Given that, I can paint 96 sq ft in 15 minutes

To find: Time taken to paint 1 square feet

From given,

96 square feet is painted in 15 minutes

Let "x" be the time taken to paint 1 square feet

Then, we can say,

96 square feet = 15 minutes

1 square feet = x minutes

This forms a proportion and we can solve the sum by cross multiplying

96 \times x = 15 \times 1\\\\x = \frac{15}{96}\\\\x = 0.15625

Thus it takes 0.15625 minutes to paint 1 square feet

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2 years ago
Solve the equation in the interval [0,2π]. If there is more than one solution write them separated by commas.
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\large\begin{array}{l} \textsf{Solve the equation for x:}\\\\ &#10;\mathsf{(tan\,x)^2+2\,tan\,x-4.76=0}\\\\\\ \textsf{Substitute}\\\\ &#10;\mathsf{tan\,x=t\qquad(t\in \mathbb{R})}\\\\\\ \textsf{so the equation &#10;becomes}\\\\ \mathsf{t^2+2t-4.76=0}\quad\Rightarrow\quad\begin{cases} &#10;\mathsf{a=1}\\\mathsf{b=2}\\\mathsf{c=-4.76} \end{cases} &#10;\end{array}


\large\begin{array}{l} \textsf{Using &#10;the quadratic formula:}\\\\ \mathsf{\Delta=b^2-4ac}\\\\ &#10;\mathsf{\Delta=2^2-4\cdot 1\cdot (-4.76)}\\\\ &#10;\mathsf{\Delta=4+19.04}\\\\ \mathsf{\Delta=23.04}\\\\ &#10;\mathsf{\Delta=\dfrac{2\,304}{100}}\\\\ &#10;\mathsf{\Delta=\dfrac{\diagup\!\!\!\! 4\cdot 576}{\diagup\!\!\!\! 4\cdot&#10; 25}}\\\\ \mathsf{\Delta=\dfrac{24^2}{5^2}} \end{array}

\large\begin{array}{l}&#10; \mathsf{\Delta=\left(\dfrac{24}{5}\right)^{\!2}}\\\\ &#10;\mathsf{\Delta=(4.8)^2}\\\\\\ &#10;\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\ &#10;\mathsf{t=\dfrac{-2\pm\sqrt{(4.8)^2}}{2\cdot 1}}\\\\ &#10;\mathsf{t=\dfrac{-2\pm 4.8}{2}}\\\\ \mathsf{t=\dfrac{\diagup\!\!\!\! &#10;2\cdot (-1\pm 2.4)}{\diagup\!\!\!\! 2}}\\\\\mathsf{t=-1\pm 2.4} &#10;\end{array}

\large\begin{array}{l} \begin{array}{rcl} &#10;\mathsf{t=-1-2.4}&~\textsf{ or }~&\mathsf{t=-1+2.4}\\\\ &#10;\mathsf{t=-3.4}&~\textsf{ or }~&\mathsf{t=1.4} \end{array} &#10;\end{array}


\large\begin{array}{l} \textsf{Both &#10;are valid values for t. Substitute back for }\mathsf{t=tan\,x:}\\\\ &#10;\begin{array}{rcl} \mathsf{tan\,x=-3.4}&~\textsf{ or &#10;}~&\mathsf{tan\,x=1.4} \end{array}\\\\\\ \textsf{Take the inverse &#10;tangent function:}\\\\ \begin{array}{rcl} &#10;\mathsf{x=tan^{-1}(-3.4)+k\cdot \pi}&~\textsf{ or &#10;}~&\mathsf{x=tan^{-1}(1.4)+k\cdot \pi}\\\\ &#10;\mathsf{x=-tan^{-1}(3.4)+k\cdot \pi}&~\textsf{ or &#10;}~&\mathsf{x=tan^{-1}(1.4)+k\cdot \pi} \end{array}\\\\\\ &#10;\textsf{where k in an integer.} \end{array}

__________


\large\begin{array}{l}&#10; \textsf{Now, restrict x values to the interval &#10;}\mathsf{[0,\,2\pi]:}\\\\ \bullet~~\textsf{For }\mathsf{k=0:}\\\\ &#10;\begin{array}{rcl} &#10;\mathsf{x=-tan^{-1}(3.4)


\large\begin{array}{l}&#10; \bullet~~\textsf{For }\mathsf{k=1:}\\\\ \begin{array}{rcl} &#10;\mathsf{x=-tan^{-1}(3.4)+\pi}&~\textsf{ or &#10;}~&\mathsf{x=tan^{-1}(1.4)+\pi} \end{array}\\\\\\ &#10;\boxed{\begin{array}{c}\mathsf{x=-tan^{-1}(3.4)+\pi} &#10;\end{array}}\textsf{ is in the 2}^{\mathsf{nd}}\textsf{ quadrant.}\\\\ &#10;\mathsf{x\approx 1.86~rad~~(106.39^\circ)}\\\\\\ &#10;\boxed{\begin{array}{c}\mathsf{x=tan^{-1}(1.4)+\pi} \end{array}}\textsf{&#10; is in the 3}^{\mathsf{rd}}\textsf{ quadrant.}\\\\ \mathsf{x\approx &#10;4.09~rad~~(234.46^\circ)}\\\\\\ \end{array}


\large\begin{array}{l}&#10; \bullet~~\textsf{For }\mathsf{k=2:}\\\\ \begin{array}{rcl} &#10;\mathsf{x=-tan^{-1}(3.4)+2\pi}&~\textsf{ or &#10;}~&\mathsf{x=tan^{-1}(1.4)+2\pi>2\pi~~\textsf{(discard)}} &#10;\end{array}\\\\\\ \boxed{\begin{array}{c}\mathsf{x=-tan^{-1}(3.4)+2\pi} &#10;\end{array}}\textsf{ is in the 4}^{\mathsf{th}}\textsf{ quadrant.}\\\\ &#10;\mathsf{x\approx 5.00~rad~~(286.39^\circ)} \end{array}


\large\begin{array}{l}&#10; \textsf{Solution set:}\\\\ &#10;\mathsf{S=\left\{tan^{-1}(1.4);\,-tan^{-1}(3.4)+\pi;\,tan^{-1}(1.4)+\pi;\,-tan^{-1}(3.4)+2\pi\right\}}&#10; \end{array}


<span>If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2071152</span>


\large\textsf{I hope it helps.}


Tags: <em>trigonometric trig quadratic equation tangent tan solve inverse symmetry parity odd function</em>

6 0
3 years ago
What is the answer to this?
Trava [24]

Answer:

<h2>C. 97.6 in³</h2>

Step-by-step explanation:

The formula of a volume of a cylinder:

V=\pi r^2H

r - radius

H - height

We have

2r=4\ in\to r=2\ in\\\\H=8\ in

Substitute:

V_1=\pi(2^2)(8)=\pi(4)(8)=32\pi\ in^3

The formula of a volume of a cone:

V=\dfrac{1}{3}\pi r^2H

We have

r=2\ in\\\\H=0.7\ in

Substitute:

V_2=\dfrac{1}{3}\pi(2^2)(0.7)=\dfrac{1}{3}\pi(4)(0.7)=\dfrac{1}{3}\pi(2.8)=\dfrac{2.8\pi}{3}=\dfrac{28\pi}{30}=\dfrac{14\pi}{15}\ in^3

The volume of the plastic object:

V=V_1-V_2

V=32\pi-\dfrac{14\pi}{15}=\dfrac{480\pi}{15}-\dfrac{14\pi}{15}=\dfrac{466\pi}{15}\\\\\pi\approx3.141\\\\V\approx\dfrac{466}{15}\cdot3.141\approx97.6

7 0
3 years ago
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