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jeka57 [31]
3 years ago
8

Consider the polynomial p(x) = x^3 + 4x^2 + 6x − 36.

Mathematics
1 answer:
Nady [450]3 years ago
4 0

Answer:

a. attached graph; zero real: 2

b. p(x) = (x - 2)(x + 3 + 3i)(x + 3 - 3i)

c. the solutions are 2, -3-3i and -3+3i

Step-by-step explanation:

p(x) = x³ + 4x² + 6x - 36

a. Through the graph, we can see that 2 is a real zero of the polynomial p. We can also use the Rational Roots Test.

p(2) = 2³ + 4.2² + 6.2 - 36 = 8 + 16 + 12 - 36 = 0

b. Now, we can use Briott-Ruffini to find the other roots and write p as a product of linear factors.

2 |  1     4     6    -36

     1      6    18     0

x² + 6x + 18 = 0

Δ = 6² - 4.1.18 = 36 - 72 = -36 = 36i²

√Δ = 6i

x = -6±6i/2 = 2(-3±3i)/2

x' = -3-3i

x" = -3+3i

p(x) = (x - 2)(x + 3 + 3i)(x + 3 - 3i)

c. the solutions are 2, -3-3i and -3+3i

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A bridge underpass in the shape of an elliptical arch, that is, half of an ellipse, is 40 feet wide and 13 feet high. An eight f
Liula [17]

Answer:

the truck should not be higher than 12.934 ft

Step-by-step explanation:

if the bridge underpass is half of an ellipse , then the arch function will be B(x,y) such that

x²/a² + y²/b² = 1  , for x≥0 and y≥0 and y=0 for x≤0

where x= width , y= height

for

a= length of the bridge underpass = 40 ft

b= height of the bridge underpass = 13 ft

x²/(40 ft)² + y²/(13 ft)² = 1

that represents all the points of the arch

the maximum height is reached when the truck touches the arch, then

for a point x= 4 ft wide , since 8 foot wide represents 4 foot to the left (x=-4) and 4 foot to the right (x=4)

then

(4 ft)²/(40 ft)² + y²/(13 ft)² = 1

1/100 - y²/(13 ft)² = 1

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y=  12.934 ft

then the truck should not be higher than 12.934 ft

7 0
3 years ago
mr. huber bought a block of fudge that weighed 2/5 of a pound. he cut the fudge into 6 equal pieces. what was the weight of each
Marina86 [1]

Answer:

\large \boxed{\frac{1}{15} \text{ lb}}

Step-by-step explanation:

You are starting with ⅖ lb of fudge. Then you cut it into six pieces.

You want to evaluate ⅖ ÷ 6

1. Treat the integer as a fraction

\dfrac{2}{5}\div{\dfrac{6}{1}}

2. Invert the divisor and change divide to multiply

\dfrac{2}{5}\times{\dfrac{1}{6}}

3. Divide numerator and denominator by 2

\dfrac{1}{5}\times{\dfrac{1}{3}}

4. Multiply numerators and denominators

\dfrac{1}{5}\times{\dfrac{1}{3}} = \mathbf{\dfrac{1}{15}}\\\\\text{Each piece of fudge weighs $\large \boxed{\mathbf{\frac{1}{15}} \textbf{ lb}}$}

7 0
4 years ago
Yea thank for the answer
kolbaska11 [484]

Answer:

Step-by-step explanation:

Hello!

10*17 = > 170 cm²

(3,14 * 8)/4 =>6,28cm²

170 + 6,28 => 176,28cm²

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