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12345 [234]
3 years ago
9

Mina, bobby, and Julia each have the same number of pencils. 2/6 of Mina's pencils are red, 2/3 of Bobby's pencils are red, and

2/4 of Julia's pencils are red. Who has more pencils, Julia or Bobby?
Mathematics
1 answer:
RUDIKE [14]3 years ago
6 0
Bobby. If you find the common denominator here which is 12 then Mina has 4/12 red pencils, bobby has 8/12 red pencils, and julia has 6/12 repent ils. Therefore bobby has the most red pencils.
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Please give 5 addition of polynomial examples​
Dafna1 [17]

Answer:

Addition of Polynomials

1Add: 3x3 – 5x2 + 8x + 10, 15x3 – 6x – 23, 9x2 – 4x + 15 and -8x3 + 2x2 – 7x.

2Add: 7a + 5b, 6a – 6b + 3c and -5a + 7b + 4c. ...

3Add: 3a2 + ab – b2, -a2 + 2ab + 3b2 and 3a2 – 10ab + 4b2 ...

4Add: 5x + 3y, 4x – 4y + z and -3x + 5y + 2z. First we need to write in the addition form. ...

Step-by-step explanation:

Hope it helps ya ItzAlex

3 0
2 years ago
Shirts are on sale for 25% of the original price of $45 what is the sale price
OverLord2011 [107]

Answer: Sale price is $11.25

Explanation:

0.25 x 45=11.25

5 0
2 years ago
Simplify 8 over negative 3 divided by negative 4 over 9 . (5 points) <br> -8<br> -6<br> 6<br> 8
denpristay [2]

Answer:

6

Step-by-step explanation:

(8/3) / (-4/9)

when dividing fractions you use the acronym k.c.f (keep change flip)

keep the first fraction

change the operation (divide to multiply)

flip the other fraction

(8/-3) x (9/4)

after re-writing your expression, you want to solve

do this by multiplying both numerators by each other and same with the denominators

(-8x9) / (3x-4)

then you get a new fraction,

-72/-12

simplify,

-72/-12 = 6 (multiplying/dividing 2 negatives will always give you a positive)

3 0
2 years ago
What is the x-intercept of the graph 4x-6y=12
kap26 [50]

Answer

x=3 or (3,0) and if you need it y=-2x+3

Step-by-step explanation:

I'm not very good at explaining things, sorry

4 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
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