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Gala2k [10]
2 years ago
11

What's cheaper, paying $20 or 20% off of $30? NEED HELP!!

Mathematics
2 answers:
Shtirlitz [24]2 years ago
6 0
$20 due to 20% off of $30 is only $24
slava [35]2 years ago
3 0

Answer:

$20 is cheaper :) (20% off of $30 would be $24!)

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Solve the division problem. Round answer to the nearest hundredth.<br><br> 9.2.15 2.0 6 3
UNO [17]

Answer:

6

Step-by-step explanation:

5 0
3 years ago
Plzzzzzzz anyone helpppppppp
sladkih [1.3K]
It's c.
for example:
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8 0
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Read 2 more answers
A scale drawing of a living room is shown below.
prisoha [69]

Answer:

A=80 square feet.

Step-by-step explanation:

First, we find the area of the model. 6*4=24. Now, due to the ratio of 1:40, we know that every inch is actually forty inches. So, you do this:

24*40=960.

So, the area of the actual living room is 960 inches. Or 80 feet.

3 0
3 years ago
The height of the Empire State Building is 1250 feet tall. Your friend, who is 75 inches tall, is standing nearby and casts a sh
Lyrx [107]
Answer:

The length of the building's shadow = 550.66 ft

Explanations:

The height of the Empie State Building = 1250 feet

The friend's height = 75 inches

The length of the friend's shadow = 33 inches

\frac{Actual\text{ height of the friend}}{\text{Length of the friend's shadow}}=\text{ }\frac{Height\text{ of the building}}{\text{Length of the building's shadow}}\begin{gathered} \frac{75}{33}=\text{ }\frac{1250}{\text{Length of the building's shadow}} \\ 2.27\text{ =  }\frac{1250}{\text{Length of the building's shadow}} \\ \text{Length of the building's shadow = }\frac{1250}{2.27} \\ \text{Length of the building's shadow = }550.66\text{ f}et \end{gathered}

6 0
1 year ago
This is a geometry question, i need something quickly :)
Marysya12 [62]

Answer:

hope it helps mark me brainlieast!

Step-by-step explanation:

<em>For triangle ABC with sides  a,b,c  labeled in the usual way, </em>

<em> </em>

<em>c2=a2+b2−2abcosC  </em>

<em> </em>

<em>We can easily solve for angle  C . </em>

<em> </em>

<em>2abcosC=a2+b2−c2  </em>

<em> </em>

<em>cosC=a2+b2−c22ab  </em>

<em> </em>

<em>C=arccosa2+b2−c22ab  </em>

<em> </em>

<em>That’s the formula for getting the angle of a triangle from its sides. </em>

<em> </em>

<em>The Law of Cosines has no exceptions and ambiguities, unlike many other trig formulas. Each possible value for a cosine maps uniquely to a triangle angle, and vice versa, a true bijection between cosines and triangle angles. Increasing cosines corresponds to smaller angles. </em>

<em> </em>

<em>−1≤cosC≤1  </em>

<em> </em>

<em>0∘≤C≤180∘  </em>

<em> </em>

<em>We needed to include the degenerate triangle angles,  0∘  and  180∘,  among the triangle angles to capture the full range of the cosine. Degenerate triangles aren’t triangles, but they do correspond to a valid configuration of three points, namely three collinear points. </em>

<em> </em>

<em>The Law of Cosines, together with  sin2θ+cos2θ=1 , is all we need to derive most of trigonometry.  C=90∘  gives the Pythagorean Theorem;  C=0  and  C=180∘  give the foundational but often unnamed Segment Addition Theorem, and the Law of Sines is in there as well, which I’ll leave for you to find, just a few steps from  cosC=  … above. (Hint: the Law of Cosines applies to all three angles in a triangle.) </em>

<em> </em>

<em>The Triangle Angle Sum Theorem,  A+B+C=180∘ , is a bit hard to tease out. Substituting the Law of Sines into the Law of Cosines we get the very cool </em>

<em> </em>

<em>2sinAsinBcosC=sin2A+sin2B−sin2C  </em>

<em> </em>

<em>Showing that’s the same as  A+B+C=180∘  is a challenge I’ll leave for you. </em>

<em> </em>

<em>In Rational Trigonometry instead of angle we use spreads, squared sines, and the squared form of the formula we just found is the Triple Spread Formula, </em>

<em> </em>

<em>4sin2Asin2B(1−sin2C)=(sin2A+sin2B−sin2C)2  </em>

<em> </em>

<em>true precisely when  ±A±B±C=180∘k , integer  k,  for some  k  and combination of signs. </em>

<em> </em>

<em>This is written in RT in an inverted notation, for triangle  abc  with vertices little  a,b,c  which we conflate with spreads  a,b,c,  </em>

<em> </em>

<em>(a+b−c)2=4ab(1−c)  </em>

<em> </em>

<em>Very tidy. It’s an often challenging third degree equation to find the spreads corresponding to angles that add to  180∘  or zero, but it’s a whole lot cleaner than the trip through the transcendental tunnel and back, which almost inevitably forces approximation.</em>

6 0
3 years ago
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