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3 years ago

4 mm

Mathematics

1 answer:

Schach [20]3 years ago

7 0

Answer:

Area = 40mm

Example:

The below-solved example problem may be useful to understand how the values are being used in the mathematical formulas to find the trapezoid area.

Example Problem :

Find the area of a trapezoid having the bottom length a = 15 cm, top length b = 12 & height h = 9 cm?

Solution :

The given values

bottom length a = 15 cm

top length b = 12 cm

height h = 10 cm

Step by step calculation

formula to find area = ((a + b)/2) h

substitute the values

= ((15 + 12)/2) x 9

= 121.5 cm2

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Answer:

I think its b

Step-by-step explanation:

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How do you solve the equation F = 9/5(K - 273.15) + 32 when you're solving for K?

Mumz [18]

$\displaystyle\\Answer:\ K=\frac{5}{9}(F-32^0)+273.15^0$

Step-by-step explanation:

$\displaystyle\\F=\frac{9}{5} (K-273.15)+32\\F-32=\frac{9}{5} (K-273.15)+32-32\\F-32=\frac{9}{5} (k-273.15)\\Multiply\ both\ parts\ of\ the\ equation\ by\ \frac{5}{9} \\(\frac{5}{9})(F-32)=\frac{9}{5} (K-273.15)(\frac{5}{9} )\\\\\frac{5}{9} (F-32)=K-273.15\\\\\frac{5}{9} (F-32)+273.15=K-273.15+273.15\\\\\frac{5}{9} (F-32)+273.15=K$

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2 years ago

The perimeter of a rectangular garden is 346. if the width of the garden is 79 feet, what is its length?

Nostrana [21]

Perimeter = L + L + W + W
346 = L + L + 79 + 79
346 = L + L + 158
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188 = L + L
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3 years ago

The bearing of Q from P is 150

Lynna [10]

Answer:

(i) see first attached diagram

(ii) QR = 22.67 km (2 dp)

(iii) 243°

Step-by-step explanation:

A bearing is the angle in degrees measured clockwise from north.

Part (i)

see first attached diagram

Part (ii)

The angles marked in blue on the second attached diagram are Consecutive Interior Angles. In this case they add to 180° as the North lines are parallel.

The sum of angles around a point is 360°

⇒ m∠QPR (shown in red on the second attached diagram) = 360 - 150 - (180 - 15) = 45°

To calculate the distance between Q and R (marked in red on the attached diagram), use the cosine rule.

⇒ QR² = PR² + QP² - 2(PR)(QP)cos(QPR)

⇒ QR² = 32² + 24² - 2(32)(24)cos(45)

⇒ QR² = 513.8839841...

⇒ QR = √513.8839841...

⇒ QR = 22.67 km (2 dp)

Part (iii)

We need to find the angle marked in green on the third attached diagram.

To do this, we need to find the angle marked in pink and the angle marked in orange, then subtract them from 360°

Pink angle = 180 - 150 = 30° (using the same consecutive angle theorem as before)

Orange angle (o) using the sine rule:

⇒ sin(o)/32 = sin(45)/QR

⇒ sin(o) = 32sin(45)/22.67

⇒ sin(o) = 0.9981652337...

⇒ o = 86.52868176...°

Therefore, the green angle = 360 - 30 - 86.52868176... = 243° (nearest degree)

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