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Nikitich [7]
3 years ago
10

If you roll 2 dice what are the odds of getting both even numbers

Mathematics
1 answer:
NNADVOKAT [17]3 years ago
4 0
It would be 6 out of 12 chances
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Determine the x-intercept for 3x + 2y = 14.
spin [16.1K]

First isolate y

3x + 2y = 14

-3x              -3x

2y = -3x + 14

y = -3/2x + 14/2

y = -3/2x + 7

X-intercept is found when y = 0

0 = -3/2x + 7

-7 = -3/2x

x = 14/3

x-in. = (14/3,0)

3 0
3 years ago
Can someone help me? In this net, the two triangles are right triangles. All quadrilaterals are rectangles. What is its surface
Nikitich [7]

Answer:

Net Area = 168

Step-by-step explanation:

See attachment for complete question

I'll make reference to the attachment, when needed.

The shapes that make up the net are labelled 1 to 5.

So, to calculate the surface area of the net, we simply calculate the surface area of each label.

Label 1: Rectangle

Area = Length * Width

Area = 10 * 5

Area = 50

Label 2: Rectangle

Area = Length * Width

Area = 8 * 5

Area = 40

Label 3: Rectangle

Area = Length * Width

Area = 6 * 5

Area = 30

Label 4: Triangle

Area = ½Base * Height

Base = 6 and Height = 8

So:

Area = ½ * 6 * 8

Area = 24

Label 5: Triangle

Area = ½Base * Height

Base = 6 and Height = 8

So:

Area = ½ * 6 * 8

Area = 24

So, the net area is the summation of the calculated areas of label 1 to 5

Net Area = 50 + 40 + 30 + 24 + 24

Net Area = 168

5 0
3 years ago
What is the unit rate for 3 boxes for $5
Andreas93 [3]
I think the answer to your question is
3
/
5
6 0
3 years 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
2 years ago
Need Help fast!!!<br> Please!
Andrej [43]
There should be a question to answer
4 0
3 years ago
Read 2 more answers
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