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zloy xaker [14]
3 years ago
9

Which of the following is the surface area of the right cylinder below radius 7 height 2

Mathematics
2 answers:
Dimas [21]3 years ago
5 0
7^2 times 2 which is 49.2 and it is 98
juin [17]3 years ago
3 0

\bf \textit{surface area of a cylinder}\\\\ SA=2\pi r(h+r)~~ \begin{cases} r=radius\\ h=height\\ -----\\ r=7\\ h=2 \end{cases}\implies SA=2\pi (7)(2+7) \\\\\\ SA=14\pi (9)\implies SA=126\pi

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Solve each equation (quadratic pattern)
dusya [7]

Answer:  x = 2

<u>Step-by-step explanation:</u>

2^{2x}-2^x-12=0\\\\\text{Let u = }2^x\\\\u^2-u-12=0\\(u-4)(u+3)=0\\\\u-4=0\quad and\quad u+3=0\\u=4\qquad and\quad u=-3\\\\\text{Substitute u with }2^x\\2^x=4\qquad and \quad 2^x=-3\\2^x=2^2\quad and\quad \text{not possible}\\\boxed{x=2}

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Answer:  x = 0

<u>Step-by-step explanation:</u>

3^{2x}+3^{x+1}-4=0\\\\3^{2x}+3^x\cdot3^1-4=0\\\\\text{Let u = }3^x\\u^2+3u-4=0\\\\(u+4)(u-1)=0\\\\u+4=0\quad and\quad u-1=0\\u=-4\qquad and\quad u=1\\\\\text{Substitute u with }3^x\\3^x=-4\qquad and\quad 3^x=1\\\text{not possible}\ and\quad 3^x=3^0\\.\qquad \qquad \qquad \qquad \boxed{x=0}

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Answer:  No Solution

<u>Step-by-step explanation:</u>

4^x+6\cdot 2^x+8=0\\\\2\cdot 2^x+6\cdot 2^x+8=0\\\\\text{Let u = }2^x\\2u+6u+8=0\\8u+8=0\\8u=-8\\u=-1\\\\\text{Substitute u with }2^x\\2^x=-1\\\text{not possible}

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Answer:  No Solution

<u>Step-by-step explanation:</u>

<u>9^x=3^x-6\\\\3\cdot 3^x=1\cdot 3^x-6\\\\\text{Let u = }3^x\\\\3u=u-6\\2u=-6\\u=-3\\\\\text{Substitute u with }3^x\\3^x=-3\\\text{not possible}</u>

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2 years ago
Last week a certain brand of hand lotion cost $3.00 for a 16-ounce bottle. this week the hand lotion is on sale for $2.40 for a
Yanka [14]
$3.00-$2.40 = $0.60
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6 0
3 years ago
Find the area of the shaded region. Round the nearest hundredth if necessary. YZ=14.2m
andreyandreev [35.5K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

1

   A_1  =  67.58 \ in^2

2

   A_2 =415.4 \ ft^2

3

   A_3  =  8.48 \ cm^2

4

  A_4 =  480.38 \ m^2

Step-by-step explanation:

Generally the area of a sector is mathematically represented as

         A =  \frac{\theta}{360} * \pi r^2

Now at r_1  = 11 in and  \theta_1 =  64^o

       A_1 =  \frac{64}{360} * 3.142  * 11^2

       A_1  =  67.58 \ in^2

Now at  r_2  = 20 ft in and  \theta_2  =  119 ^o

       A_2 =  \frac{119}{360} * 3.142 *  20^2

       A_2 =415.4 \ ft^2

Now at  r_3  = 6.5 cm   and  \theta_3 =  23 ^o

     A_3  =  \frac{23}{360} * 3.142 *  6.5 ^2

      A_3  =  8.48 \ cm^2

Now at  r_4  = 14.2 m   and  \theta_4 = 360 -87 =  273 ^o

         A_4 =  \frac{273}{360}  * 3.142 * 14.2^2

          A_4 =  480.38 \ m^2

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