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kvasek [131]
2 years ago
8

Find the surface area of the cylinder to the nearest tenth of a square unit. use 3.14 for pi

Mathematics
1 answer:
Alborosie2 years ago
6 0

Given:

  • Height of cylinder is 18.2 cm
  • Radius of cylinder is 3 cm

~

To Find?

  • Surface area

~

Solution:

Using formula:

  • Surface area of cylinder = 2πr (h + r)

~

Substituting values in the formula:

\longrightarrow \sf 2 × 3.14 × 3 (18.2 + 3)

\longrightarrow \sf 6.28 × 3 (21.2)

\longrightarrow \sf 18.84 × 21.2

\longrightarrow \sf 399.4 cm²

~

  • Hence, (B) 399.4 cm² is right answer.
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Identify the sine function with the given attributes.
Thepotemich [5.8K]

Answer:

<em>Last option </em>

f(t)=sin(\pi(t -3)) + 4

Step-by-step explanation:

The general sine function has the following form  

y = Asin(bt-\phi) + k

Where A is the amplitude: half the vertical distance between the highest peak and the lowest peak of the wave.  

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k is the vertical displacement.

\phi phase shift

We know that:

amplitude: 1; period: 2; phase shift: 3; vertical shift: 4

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4 0
3 years ago
Harry is trying to solve the equation 0 = 2x2 − x − 6 using the quadratic formula. He has made an error in one of the steps belo
bija089 [108]
<h3><u>Given</u> - </h3>

➙ a quadratic equation in which Harry lagged due to an error made by him, 2x² - x - 6= 0

<h3><u>To solve</u> - </h3>

➙ the given quadratic equation.

<h3><u>Concept applied</u> - </h3>

➙ We will apply the quadratic formula as given in the question. So, let's study about quadratic equation first because we are supposed to apply the formula in equation.

What is quadratic equation?

➙ A quadratic equation in the variable x is an equation of the form ax² + bx + c = 0, where a, b, c are real numbers, a ≠ 0.

Now, what is quadratic formula?

➙The roots of a quadratic equation ax + bx + c = 0 are given by \sf{\:\frac{-b \pm\: \sqrt {b ^ 2 - 4ac}}{2a}} provided b - 4ac ≥ 0.

<h3><u>Solution</u> - </h3>

here as per the given quadratic equation,

a = 2, b = -1 and c = -6

putting in the formula,

\implies\sf{x=\frac{-(-1) \pm\: \sqrt {(-1)^2 - 4(2)(-6)}}{2(2)}}

\implies\sf{x=\frac{1 \pm\: \sqrt {1+48}}{4}}

\implies\sf{x=\frac{1 \pm\: \sqrt {49}}{4}}

\implies\sf{x=\frac{1 \pm\: 7}{4}}

Solving one by one,

\implies\sf{x=\frac{1 + \: 7}{4}}

\implies\sf{x=\frac{8}{4}}

\implies{\boxed{\bf{x=2}}}

________________

\implies\sf{x=\frac{1 - \: 7}{4}}

\implies\sf{x=\frac{-6}{4}}

\implies{\boxed{\bf{x=\frac{-3}{2}}}}

________________________________

<em><u>Note</u> - Hey dear user!! You haven't provided the solution which was solved by Harry (A.T.Q). Please go through the solution as it will help you to find the error done by Harry.</em>

<em>________________________________</em>

Hope it helps!! (:

4 0
2 years ago
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