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3 years ago

8

The polynomial function F(x)= 2x^2 +4 has a critical point at which of the

1 answer:

Whitepunk [10]3 years ago

8 0

It will be (C) because you have to find the GCF first than simplify it and it gives you 2

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At Wendy’s, 3 Baconators and 2 Frosty’s cost $21.50. If 5 Baconator’s and 1 Frosty cost $30, find the cost of one Baconator and

Verizon [17]

Suggartits it’s easy the answer is 3Baconatore 2 frosty free points bby

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3 years ago

One number is 2 more than another number. The product of the numbers is 440. Find the numbers.

Bumek [7]

Answer: 20 and 22

The product of 20 and 22 is 440 and 22 is 2 more than 20.

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2 years ago

Read 2 more answers

The driving distance for the top 100 golfers on the PGA tour is between 284.7 and 310.6 yards (according to GolfWeek). Assume th

Usimov [2.4K]

Answer:

a) $f(x) = \frac{1}{25.9}$

b) 0.2046 = 20.46% probability the driving distance for one of these golfers is less than 290 yards

Step-by-step explanation:

Uniform probability distribution:

An uniform distribution has two bounds, a and b.

The probability of finding a value of at lower than x is:

$P(X < x) = \frac{x - a}{b - a}$

The probability of finding a value between c and d is:

$P(c \leq X \leq d) = \frac{d - c}{b - a}$

The probability of finding a value above x is:

$P(X > x) = \frac{b - x}{b - a}$

The probability density function of the uniform distribution is:

$f(x) = \frac{1}{b-a}$

The driving distance for the top 100 golfers on the PGA tour is between 284.7 and 310.6 yards.

This means that $a = 284.7, b = 310.6$ .

a. Give a mathematical expression for the probability density function of driving distance.

$f(x) = \frac{1}{b-a} = \frac{1}{310.6-284.7} = \frac{1}{25.9}$

b. What is the probability the driving distance for one of these golfers is less than 290 yards?

$P(X < 290) = \frac{290 - 284.7}{310.6-284.7} = 0.2046$

0.2046 = 20.46% probability the driving distance for one of these golfers is less than 290 yards

3 0

3 years ago

The temperature function (in degrees Fahrenheit) in a three dimensional space is given by T(x, y, z) = 3x + 6y - 6z + 1. A bee i

madam [21]

You're looking for the extreme values of $T(x,y,z)=3x+6y-6z+1$ subject to $x^2+y^2+z^2=9$ . The Lagrangian is

$L(x,y,z,\lambda)=3x+6y-6z+1+\lambda(x^2+y^2+z^2-9)$

with critical wherever the partial derivatives vanish:

$L_x=3+2\lambda x=0\implies x=-\dfrac3{2\lambda}$

$L_y=6+2\lambda y=0\implies y=-\dfrac3\lambda$

$L_z=-6+2\lambda z=0\implies z=\dfrac3\lambda$

$L_\lambda=x^2+y^2+z^2-9=0$

Substituting the first three solutions into the last equation gives

$\dfrac9{4\lambda^2}+\dfrac9{\lambda^2}+\dfrac9{\lambda^2}=9$

$\implies\lambda=\pm\dfrac32$

$\implies x=1,y=2,z=-2\text{ or }x=-1,y=-2,z=2$

At these points, we have

$T(1,2,-2)=28$

$T(-1,-2,2)=-26$

so the highest temperature the bee can experience is 28º F at the point (1, 2, -2), and the lowest is -26º F at the point (-1, -2, 2).

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3 years ago

PLS help with this immediately no links or guesses

shusha [124]

Answer:

B all real numbers greater than 0

3 0

3 years ago