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3 years ago

Solve The Geometry Question In The Image

Mathematics

2 answers:

Readme [11.4K]3 years ago

4 0

By the Pythagorean theorem:

 $AD= \sqrt{AB^2-BD^2}= \sqrt{9.4^2-7.9^2}= \sqrt{25.95} \approx 5.09 \ units\\\\ CD= \sqrt{BC^2-BD^2}= \sqrt{23.2^2-7.9^2}= \sqrt{475.83} \approx 21.81 \ units \\ \\ AC=AD+CD=5.09+21.81=26.9 \ units$

kondaur [170]3 years ago

3 0

use the pythagorean theorem twice
alrighty
so
remember for legs length a and b and hytponuse c in a right triangle
a²+b²=c²

we need AD and DC

so
AD²+7.9²=9.4²
AD=√25.95

DC²+7.9²=23.2²
DC=√475.83

so
AD+DC=base=(√25.95)+(√475.83)≈26.9076

AC≈26.91 units

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What is the shape of the cross-section formed when a plane containing line AC and line EH intersects this cube? What is the area

velikii [3]

Answer:

Cross section is a rectangle.

Area of cross section = 16 sqrt(2) = 22.63 sq. units (to 2 decimals)

Step-by-step explanation:

Given a cube.

top face ABCD is parallel and congruent to bottom face EFGH ........(1)

justified by the properties of cubes

Sides AE and CH are perpendicular to faces ABCD and EFGH ..........(2)

justified by the properties of cubes

Diagonals AC and EH are congruent ......................(3)

justified by (1), congruent top and bottom faces

Consider cross-section ACHE

AC is congruent and parallel to EH (1) & (3)

EA & HC are perpendicular to AC (2)

Therefore the quadrilateral ACHE is a rectangle.

Length of diagonal AC = sqrt(4^2+4^2) = 4 sqrt(2) ..........pythagoras theorem

AE = CH = DG = 4 properties of cube, all sides equal

Area of ACHE = 4* 4sqrt(2) = 16 sqrt(2) = 22.63 sq. units

4 0

2 years ago

Solve (10b+7b^2-6b^3)-(12b^2-3b^3)

MrRa [10]

Answer:

-3b³ - 5b² + 10b

General Formulas and Concepts:

Pre-Algebra

Distributive Property

Algebra I

Combining Like Terms

Step-by-step explanation:

Step 1: Define Expression

(10b + 7b² - 6b³) - (12b² - 3b³)

Step 2: Simplify

Distribute negative: 10b + 7b² - 6b³ - 12b² + 3b³
Combine like terms (b³): -3b³ + 10b + 7b² - 12b²
Combine like terms (b²): -3b³ - 5b² + 10b

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2 years ago

Each of 16 students measured the circumference of a tennis ball by four different methods, which were: A: Estimate the circumfer

almond37 [142]

Answer:

Following are the solution to the given equation:

Step-by-step explanation:

Please find the complete question in the attachment file.

In point a:

$\to \mu=\frac{\sum xi}{n}$

$=22.8$

$\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}$

$=\sqrt{\frac{119.18}{16-1}}\\\\ =\sqrt{\frac{119.18}{15}}\\\\ = \sqrt{7.94533333}\\\\=2.8187$

In point b:

$\to \mu=\frac{\sum xi}{n}$

$=20.6875$

$\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}$

$=\sqrt{\frac{26.3375}{16-1}}\\\\=\sqrt{\frac{26.3375}{15}}\\\\ =\sqrt{1.75583333}\\\\ =1.3251$

In point c:

$\to \mu=\frac{\sum xi}{n}$

$=21$

$\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}$

$=\sqrt{\frac{2.62}{16-1}}\\\\ =\sqrt{\frac{2.62}{15}} \\\\= \sqrt{0.174666667}\\\\=0.4179$

In point d:

$\to \mu=\frac{\sum xi}{n}$

$=20.8375$

$\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}$

$=\sqrt{\frac{8.2975}{16-1}}\\\\ =\sqrt{\frac{8.2975}{15}} \\\\ =\sqrt{0.553166667} \\\\ =0.7438$

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2 years ago

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levacccp [35]

Answer:3/4 inch

Step-by-step explanation:

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Sphinxa [80]

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2 years ago