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3 years ago

Brian has 2 fish ponds in his backyard connected by a waterfall The top pond

Mathematics

1 answer:

Andrews [41]3 years ago

5 0

Answer:

182 gallons of water

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How do I solve this?

Norma-Jean [14]

Answer:

Step-by-step explanation:

Base = 25 ft

height = 8 ft

Area of triangle = $\frac{1}{2}*base * height\\\\$

$= \frac{1}{2}*25 *8\\\\= 25 * 4\\\\= 100 ft^{2}$

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2 years ago

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Help! will appreciate

Lera25 [3.4K]

Sorry I've got on idea

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2 years ago

A town has 300 residents in total, of which 5/6 are white/Caucasian Americans. How many white/Caucasian Americans reside in this

Neko [114]

Answer:

250

Step-by-step explanation:

300* 5/6 = 250

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3 years ago

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TOAD is a quadrilateral with vertices (9, 10), (18, 10), (14, 5), and (5, 5), respectively. The diagonals intersect at Point S.

klasskru [66]

A. parallelogram
B. I believe it'd be around (11.2,7.5)
C. √(10-5)^2+(9-4)^2 = 21, so yes they bisect

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2 years ago

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Given the following trigonometric ratio, enumerate the meaning ratio

Likurg_2 [28]

Answer:

The trigonometric ratios are presented below:

$\sin \theta = \frac{AC}{\sqrt{AC^{2} + BC^{2}}}$

$\cos \theta = \frac{BC}{\sqrt{AC^{2} + BC^{2}}}$

$\cot \theta = \frac{BC}{AC}$

$\sec \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{BC}$

$\csc \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{AC}$

Step-by-step explanation:

From Trigonometry we know the following definitions for each trigonometric ratio:

Sine

$\sin \theta = \frac{y}{h}$ (1)

Cosine

$\cos \theta = \frac{x}{h}$ (2)

Tangent

$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{y}{x}$ (3)

Cotangent

$\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{x}{y}$ (4)

Secant

$\sec \theta = \frac{1}{\cos \theta} = \frac{h}{x}$ (5)

Cosecant

$\csc \theta = \frac{1}{\sin \theta} = \frac{h}{y}$ (6)

Where:

$x$ - Adjacent leg.

$y$ - Opposite leg.

$h$ - Hypotenuse.

The length of the hypotenuse is determined by the Pythagorean Theorem:

$h = \sqrt{x^{2}+y^{2}}$

If $y = AC$ and $x = BC$ , then the trigonometric ratios are presented below:

$\sin \theta = \frac{AC}{\sqrt{AC^{2} + BC^{2}}}$

$\cos \theta = \frac{BC}{\sqrt{AC^{2} + BC^{2}}}$

$\cot \theta = \frac{BC}{AC}$

$\sec \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{BC}$

$\csc \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{AC}$

6 0

3 years ago