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2 years ago

Q4. The diagram shows a triangle inside a rectangle.

Mathematics

1 answer:

Yanka [14]2 years ago

4 0

The prove to show that the area of the shaded portion is 18x - 30 is as follows: 3x² + 13x - 30 - 3x² + 5x = 18x - 30

<h3>How to find the area of the shaded region?</h3>

The area of the shaded region can be represented as follows:

area of the shaded region = area of rectangle - area of triangle

Therefore,

area of the rectangle = (x + 6)(3x - 5)

area of the rectangle = 3x² - 5x + 18x - 30

area of the rectangle = 3x² + 13x - 30

area of triangle = 1 / 2 × 2x × 3x - 5

area of triangle = 3x² - 5x

Therefore,

area of the shaded portion = 3x² + 13x - 30 - 3x² + 5x

area of the shaded portion = 18x - 30

learn more on area here: brainly.com/question/21208569

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What is - 1/8 as a terminating decimal

wolverine [178]

Answer:

1/8 as a terminating decimal is 0.125

8 0

3 years ago

HELP NEEDED PLEASE!!!

postnew [5]

Using vector concepts, it is found that:

The component form is of approximately (-9.58, 7,22). It means that the ship is about 9.58 miles to the west and about 7.22 miles to the north of where the ship left the port.

<h3>How can a vector be represented in component notation?</h3>

Given a magnitude M and angle $\theta$ , then a vector V can be represented as follows in component notation:

$V = (M\cos{\theta}, M\sin{\theta})$

In this problem, the magnitude and the angle are given, respectively, by:

$M = 12, \theta = 143^\circ$

Hence:

V = [12cos(143º), 12sin(143º)] = (-9.58, 7,22).

Which means a displacement of 9.58 miles to the west(negative x = west) and 7.22 miles to the north(positive y = north).

The component form is of approximately (-9.58, 7,22). It means that the ship is about 9.58 miles to the west and about 7.22 miles to the north of where the ship left the port.

More can be learned about vectors at brainly.com/question/24606590

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8 0

2 years ago

If you are given the following stem and leaf display and asked to construct a frequency distribution chart, what would be the wi

Nuetrik [128]

Answer:

Width of intervals: 8

Step-by-step explanation:

We first look at how data is represented in a stem-leaf diagram.

Any number of the left (before -) is the stem and all numbers on right (after -) are the leaves. Each combination of stem and leaf represents one number. For example: 1 - 332 represents: 13, 13, 12.

Our data is as follows:

13, 13, 12, 24, 25, 31, 31, 35, 37, 42, 43, 41, 52, 51, 51, 52

To calculate the width of the frequency distribution chart, we have the following formula:

$Class\ width = \frac{Range}{Number\ of\ classes}$

The range of any data set = Maximum value in the data set - Minimum value in the data set

Maximum value in this case as seen from the data is 52 and minimum is 12.

Range = 52 - 12 = 40

Since we had only 5 stems in the data, we shall use that as the number of classes required in the frequency distribution chart.

$Class\ width = \frac{40}{5} = 8$

Hence, the class width in this data set will be 8.

To make the intervals, we begin from the minimum value and add 8 to it. The intervals will be:

12 - 20

20 - 28

28 - 36

36 - 44

44 - 52

Observe, that all the values of the stem lie within each interval.

For example, there are 3 values for stem 1: 12, 13, 13 and each lie in the first interval 12 - 20.

Next, the values of stem 2 are 24 and 25. Each of these value lie in the second interval 20 - 28; and henceforth.

8 0

3 years ago

Need help on number 12....

Vaselesa [24]

Answer:

the answer is A. Elijah is correct.

5 0

4 years ago

Read 2 more answers

If anyone knows about definite integrals for calculus then please I request help! I

kicyunya [14]

Answer:

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

General Formulas and Concepts:

<u>Calculus</u>

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

U-Substitution

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx$

<u>Step 2: Integrate Pt. 1</u>

<em>Identify variables for u-substitution.</em>

Set <em>u</em>: $\displaystyle u = 4x^{-2}$
[<em>u</em>] Differentiate [Basic Power Rule, Derivative Properties]: $\displaystyle du = \frac{-8}{x^3} \ dx$
[Bounds] Switch: $\displaystyle \left \{ {{x = 9 ,\ u = 4(9)^{-2} = \frac{4}{81}} \atop {x = 5 ,\ u = 4(5)^{-2} = \frac{4}{25}}} \right.$

<u>Step 3: Integrate Pt. 2</u>

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^9_5 {\frac{-8}{x^3}e^\big{4x^{-2}}} \, dx$
[Integral] U-Substitution: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{4}{81}}_{\frac{4}{25}} {e^\big{u}} \, du$
[Integral] Exponential Integration: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}(e^\big{u}) \bigg| \limits^{\frac{4}{81}}_{\frac{4}{25}}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8} \bigg( e^\Big{\frac{4}{81}} - e^\Big{\frac{4}{25}} \bigg)$
Simplify: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

4 0

2 years ago