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2 years ago

Which of the following represents the zeros of f(x) = 6x3 − 31x2 + 4x + 5?

Mathematics

1 answer:

kotegsom [21]2 years ago

8 0

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In triangle abc, the angle c is six times as large as angle

mote1985 [20]

The interior angles of a triangle add up to 180 degrees

a = 1st angle, b = 2nd angle, c = 3rd angle

c = 6a
b = a + 60
a + b + c = 180

a + (a + 60) + (6a) = 180
8a + 60 = 180
8a = 180 - 60
8a = 120
a = 120/8
a = 15 <=== here is one angle

b = a + 60
b = 15 + 60
b = 75 <=== and another one

c = 6a
c = 6(15)
c = 90 <=== and the last angle

3 0

3 years ago

A family will travel 350 miles from their house in order to reach Dallas, TX. Which inequality can be used to find all possible

Flura [38]

Tje answer is c t < or equal to (350r)

5 0

3 years ago

How many sides does a quadrillateral have

masya89 [10]

Answer: 4

Step-by-step explanation:In geometry a quadrilateral is a four-sided polygon, having four edges and four corners. The word is derived from the Latin words quadri, a variant of four, and latus, meaning side

3 0

2 years ago

ANSWER CORRET AND ILL MARK U BRAINLIEST

Kipish [7]

Answer:

two is the last one, three us the first one, and four is the second one, and six is the definition of parallelogram

6 0

3 years ago

Cos^2x+cos^2(120°+x)+cos^2(120°-x)<br>i need this asap. pls help me

o-na [289]

Answer:

$\frac{3}{2}$

Step-by-step explanation:

Using the addition formulae for cosine

cos(x ± y) = cosxcosy ∓ sinxsiny

---------------------------------------------------------------

cos(120 + x) = cos120cosx - sin120sinx

= - cos60cosx - sin60sinx

= - $\frac{1}{2}$ cosx - $\frac{\sqrt{3} }{2}$ sinx

squaring to obtain cos² (120 + x)

= $\frac{1}{4}$ cos²x + $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

--------------------------------------------------------------------

cos(120 - x) = cos120cosx + sin120sinx

= -cos60cosx + sin60sinx

= - $\frac{1}{2}$ cosx + $\frac{\sqrt{3} }{2}$ sinx

squaring to obtain cos²(120 - x)

= $\frac{1}{4}$ cos²x - $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

--------------------------------------------------------------------------

Putting it all together

cos²x + $\frac{1}{4}$ cos²x + $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x + $\frac{1}{4}$ cos²x - $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

= cos²x + $\frac{1}{2}$ cos²x + $\frac{3}{2}$ sin²x

= $\frac{3}{2}$ cos²x + $\frac{3}{2}$ sin²x

= $\frac{3}{2}$ (cos²x + sin²x) = $\frac{3}{2}$

5 0

3 years ago