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sineoko [7]
3 years ago
14

The area of a trapezoid is 243cm2. The height is 18cm and the length of one of the parallel sides is 10cm. Find the length of th

e second parallel side. Express your answer as a simplified fraction or a decimal rounded to two places.
Mathematics
1 answer:
Irina18 [472]3 years ago
8 0

Answer:

Side_2 = 17\ cm

Step-by-step explanation:

Given

Shape: Trapezoid

Area = 243cm^2

Height = 18cm

Side_1 = 10cm

Required

Determine the length of the second parallel side

The area of a trapezoid is:

Area = \frac{1}{2}(Side_1 + Side_2) * Height

Substitute values for Area, Height and Side1

243 = \frac{1}{2}(10 + Side_2) * 18

Multiply both sides by 2

2 * 243 = 2 * \frac{1}{2}(10 + Side_2) * 18

486 = (10 + Side_2) * 18

Divide both sides by 18

27 = 10 + Side_2

Side_2 = 27 - 10

Side_2 = 17\ cm

Hence;

<em>The length of the second parallel side is 17cm</em>

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Eduardwww [97]

Answer:

40% or 0.4

Step-by-step explanation:

The optimal capital structure (OCS) of a firm is defined as "the proportion of debt and equity that results in the lowest weighted average cost of capital (WACC) for the firm"

The brief explanation of this is that OCS is the factor used by a company in maximising their stock price, and this generally calls for a Debt-to-capital or "Debit-to-equity" ratio.

From the table above, the company's stock ratio is highest or maximised at 37.75 (under Projected Stock Price Column)

This can be traced to 40% under Debt/Capital ratio column

Hence, the Debt/Capital Ratio of 40%,

Because it must equate to 100%, we say that the firm's optimal capital structure is 40% debt and 60% equity.

This is also the debt to capital ratio, where the firms WACC is minimized.

5 0
4 years ago
Justin Verlander of the Astros threw a pitch 100 miles per hour. How many seconds did it take to get to home plate. The plate is
castortr0y [4]

It will take 0.4 seconds to get to the home plate.

Given:

The speed of the pitch thrown by Justin is 100 miles per hour

The distance of home plate from the pitching rubber is 60.5 feet

To find:

The time taken by a pitch to reach the home plate

Solution:

The distance home plate from the pitching rubber = d = 60.5 feet

1 mile = 5280 feet\\1 foot=\frac{1}{5280} mile\\60.5 feet=60.5 \times \frac{1}{5280} mile=\frac{60.5}{5280}miles

The speed of pitch thrown by Justin = s = 100 miles/hr

The time taken by a pitch to reach the home plate = t =?

The speed of the moving object is given by dividing the distance covered from the time taken to cover that distance.

Speed=\frac{Distance}{Time}\\s=\frac{d}{t}\\100 miles/hr=\frac{\frac{60.5}{5280}miles}{t}\\t=\frac{60.5}{5280}miles\times \frac{1}{100 miles/hr}\\= 0.000114583 hr

In an hour there are 3600 seconds, then in 0.000114583 hours will be:

1 hour=3600 seconds\\=0.000114583 \times 3600=0.4124988 s\approx 0.4 s

It will take 0.4 seconds to get to the home plate.

Learn more about conversions here:

brainly.com/question/24530464

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3 years ago
Solve the equation uding the most direct method: 3x(x+6)=-10?​
Tanzania [10]

To solve this problem, you will use the distributive property to create an equation that can be rearranged and solved using the quadratic formula.

<h3>Distribute</h3>

Use the distributive property to distribute 3x into the term (x + 6):

3x(x+6)=-10

3x^2+18x=-10

<h3>Rearrange</h3>

To create a quadratic equation, add 10 to both sides of the equation:

3x^2+18x+10=-10+10

3x^2+18x+10=0

<h3>Use the Quadratic Formula</h3>

The quadratic formula is defined as:

\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

The model of a quadratic equation is defined as ax² + bx + c = 0. This can be related to our equation.

Therefore:

  • a = 3
  • b = 18
  • c = 10

Set up the quadratic formula:

\displaystyle x=\frac{-18 \pm \sqrt{(18)^2 - 4(3)(10)}}{2(3)}

Simplify by using BPEMDAS, which is an acronym for the order of operations:

Brackets

Parentheses

Exponents

Multiplication

Division

Addition

Subtraction

Use BPEMDAS:

\displaystyle x=\frac{-18 \pm \sqrt{324 - 120}}{6}

Simplify the radicand:

\displaystyle x=\frac{-18 \pm \sqrt{204}}{6}

Create a factor tree for 204:

204 - 1, 2, 3, 4, 6, 12, 17, 34, 51, 68, 102 and 204.

The largest factor group that creates a perfect square is 4 × 51. Therefore, turn 204 into 4 × 51:

\sqrt{4\times51}

Then, using the Product Property of Square Roots, break this into two radicands:

\sqrt{4} \times \sqrt{51}

Since 4 is a perfect square, it can be evaluated:

2 \times \sqrt{51}

To simplify further for easier reading, remove the multiplication symbol:

2\sqrt{51}

Then, substitute for the quadratic formula:

\displaystyle x=\frac{-18 \pm 2\sqrt{51}}{6}

This gives us a combined root, which we should separate to make things easier on ourselves.

<h3>Separate the Roots</h3>

Separate the roots at the plus-minus symbol:

\displaystyle x=\frac{-18 + 2\sqrt{51}}{6}

\displaystyle x=\frac{-18 - 2\sqrt{51}}{6}

Then, simplify the numerator of the roots by factoring 2 out:

\displaystyle x=\frac{2(-9 + \sqrt{51})}{6}

\displaystyle x=\frac{2(-9 - \sqrt{51})}{6}

Then, simplify the fraction by reducing 2/6 to 1/3:

\boxed{\displaystyle x=\frac{-9 + \sqrt{51}}{3}}

\boxed{\displaystyle x=\frac{-9 - \sqrt{51}}{3}}

The final answer to this problem is:

\displaystyle x=\frac{-9 + \sqrt{51}}{3}

\displaystyle x=\frac{-9 - \sqrt{51}}{3}

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Answer:

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Step-by-step explanation:

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