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Paha777 [63]
3 years ago
8

Subtract the quotient of 15 ÷ 3 from the sum of 33 and 7.

Mathematics
2 answers:
Tems11 [23]3 years ago
6 0
15:3=5\\33+7=40\\\\33+7-15:3=40-5=\fbox{35}\leftarrow answer\ \boxed{C}
lys-0071 [83]3 years ago
4 0
In other words, subtract 5 from 40 = the answer is 35 (c)
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at the end of his shift, Bruno had $340 worth of tips all in $10 and $20 bills .if he had two more $20 bills than $10 bill .how
PSYCHO15rus [73]
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8 0
3 years ago
D 2.<br> The sum of 3x2 – 8x – 2 and 4x – 2 is
maw [93]

Answer:

3x^2 -4x -4

Step-by-step explanation:

3x^2 – 8x – 2 + 4x – 2

Combine like terms

3x^2 -8x+4x -2 -2

3x^2 -4x -4

8 0
3 years ago
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Please help me this is my last question
soldier1979 [14.2K]

Answer:

x=-22

Step-by-step explanation:

6 0
2 years ago
Solve the equation in the interval [0,2π]. If there is more than one solution write them separated by commas.
Sedaia [141]
\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
Find the volume of a cylinder that has a diameter of 12 in. and a height of 15 in
daser333 [38]

Answer:

<h3>\sf{ \boxed{ \bold{1697.14}}}</h3>

Step-by-step explanation:

Given,

diameter ( d ) = 12 in

height ( h ) = 15 in

<u>finding </u><u>the </u><u>radius </u><u>of </u><u>a </u><u>cylinder</u>

Radius is just half of diameter.

Radius ( r ) = 12 / 2 = 6 in

<u>finding </u><u>the </u><u>volume </u><u>of </u><u>a </u><u>cylinder </u><u>having </u><u>radius </u><u>of </u><u>6</u><u> </u><u>in </u><u>and </u><u>height </u><u>of </u><u>1</u><u>5</u><u> </u><u>in</u>

Volume of a cylinder = <u>\sf{\pi  \:  {r}^{2} h}</u>

⇒\sf{ \frac{22}{7}  \times  {6}^{2}  \times 15}

⇒\sf{ \frac{22}{7}  \times  36 \times 15}

⇒<u>\sf{1697.14} \: in</u>

Hope I helped!

Best regards!!

6 0
3 years ago
Read 2 more answers
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