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myrzilka [38]
1 year ago
5

Kenneth drew a scale drawing of an arcade an air hockey table Wich is 10 feet long In real life is 5 inches long in the drawing

Wich scale did Kenneth use?
Mathematics
1 answer:
s344n2d4d5 [400]1 year ago
5 0

The scale of the drawing is 2 feet : 1 inch

<h3>How to determine the scale of the drawing?</h3>

The given parameters are:

Scale measurement = 10 feet

Real life = 5 inches

The scale of the drawing is represented as;

Scale = Scale measurement : Real life

So, we have

Scale = 10 feet : 5 inches

Simplify

Scale = 2 feet : 1 inch

Hence, the scale of the drawing is 2 feet : 1 inch

Read more about scale at:

brainly.com/question/15891755

#SPJ1

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Hi there,

Properties of Equality is two equations that have the same solution which are called equivalent equations. For example: 5+3=2+6 which all equal to 8.

In geometry, properties of Equality are the reflexive (2x=2x, a=a, b+1=b+1), symmetric (2a=3b then 3b=2a) and transitive property (if a+1=3 and 3=1-b, then a+1=1-b)

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7 0
3 years ago
Three side lengths of a right triangle are given which side length should you substitute for the hypotenuse in Pythagorean theor
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Answer:

The longest length which you are given.

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3 years ago
Of the 21 children at a park, 5 of them are playing on the swings. What is the approximate
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The Answer Is B

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3 0
2 years ago
Find the exact values of a) sec of theta b)tan of theta if cos of theta= -4/5 and sin&lt;0
Gre4nikov [31]

Answer:

Using trigonometric ratio:

\sec \theta = \frac{1}{\cos \theta}

\tan \theta = \frac{\sin \theta}{\cos \theta}

From the given statement:

\cos \theta = -\frac{4}{5} and sin < 0

⇒\theta lies in the 3rd quadrant.

then;

\sec \theta = \frac{1}{-\frac{4}{5}} = -\frac{5}{4}

Using trigonometry identities:

\sin \theta = \pm \sqrt{1-\cos^2 \theta}

Substitute the given values we have;

\sin \theta = \pm\sqrt{1-(\frac{-4}{5})^2 } =\pm\sqrt{1-\frac{16}{25}} =\pm\sqrt{\frac{25-16}{25}} =\pm \sqrt{\frac{9}{25} } = \pm\frac{3}{5}

Since, sin < 0

⇒\sin \theta = -\frac{3}{5}

now, find \tan \theta:

\tan \theta = \frac{\sin \theta}{\cos \theta}

Substitute the given values we have;

\tan \theta = \frac{-\frac{3}{5} }{-\frac{4}{5} } = \frac{3}{5}\times \frac{5}{4} = \frac{3}{4}

Therefore, the exact value of:

(a)

\sec \theta =-\frac{5}{4}

(b)

\tan \theta= \frac{3}{4}

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3 years ago
Simplify (8x^4)(4x^3) divided by 7x^7
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