Please help me 20 points

Answers

Answer 1

Answer:

Answer:

sureeeeeeeeeeeeeeeeee

Step-by-step explanation:

Related Questions

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Consider the figure.Find AB if BC = 4, BD = 5, and AD = 5

WILL BE MARKED BRAINLYESTRay OJ bisects Angle LOK.

Measurement of Angle LOK = 82 degrees.

Solve for x.

Answers

Answer:

x = 14

Step-by-step explanation:

<LOJ = 3x

<KOJ = (2x + 12)°

<LOK = 82°

m<LOJ + m<KOJ = m<LOK (angle addition postulate)

3x + 2x + 12 = 82 (substitution)

5x + 12 = 82

5x = 82 - 12 (Subtraction property of equality)

5x = 70

x = 70/5 (division property of equality)

x = 14

The equation T^2=A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A, in astronomical units, AU. If planet Y is twice the mean distance from the sun as planet X, by what factor is the orbital period increased?

Answers

the answer for your question is going to be. 2^((3)/(2)) I know this cause I just took a quiz with this question on it

Answer:

2 Superscript three-halves

Step-by-step explanation:

Edge

Jerry is a judge. He hears 555 cases every 2 3/8. Jerry hears cases at a constant rate. How many cases does he hear per hour?

Answers

About 2.1 cases so it maybe 2 full cases. Hope this helped!

Simplify The square root of 5 (6-4 the square root of 3)

Answers

Answer:

7.75

Step-by-step explanation:

6-4=2

2 times the square root of 3=3.46410161514

square root of 5 times 3.46410161514=7.74596669242

to 2dp=7.75

Giving a test to a group of students, the grades and gender are summarized below. Round your answers to 4 decimal places. A B C Total
Male 16 13 6 35
Female 2 3 8 13
Total 18 16 14 48
If one student is chosen at random:
1. Find the probability that the student was female AND got a "B".
2. Find the probability that the student was male AND got a "A".
3. Find the probability that the student got a B.

Answers

Answer:

1). 0.1667

2). 0.3333

3). 0.3333

Step-by-step explanation:

1). Probability that the student was female and got a 'B'

= $\frac{\text{Number of female students who got B}}{\text{Total number of students}}$

= $(3)/(48)=(1)/(16)$

= 0.1667

2). Probability that the student was male and got an 'A'

= $\frac{\text{Number of male students who got A }}{\text{Total number of students}}$

= $(16)/(48)$

= $(1)/(3)$

= 0.3333

3). Probability that the student got a B = $\frac{\text{Number of students who got B}}{\text{Total number of students}}$

= $(16)/(48)$

= $(1)/(3)$

= 0.3333

Find the critical points, domain endpoints, and local extreme values for the functiony=x^2/5(x+3)

a. What is/are the critical point(s) and domain endpoint(s) where f' is undefined?
b. What is/are the critical point(s) and domain endpoint(s) where f' is 0?
c. From the critical point(s) and domain endpoint(s), what is/are the points corresponding to local maxima?
d. From the critical point(s) and domain endpoint(s), what is/are the points corresponding to local minima?

Answers

Answer:

a) $x = -3$ , b) $x = 0$ , $x = -6$ , c) $x = 0$ , d) $x = -6$

Step-by-step explanation:

a) Let derive the function:

$f'(x) = (10\cdot x \cdot (x+3)-5\cdot x^(2))/(25\cdot (x+3)^(2))$

$f'(x)$ is undefined when denominator equates to zero. The critical point is:

$x = -3$

b) $f'(x) = 0$ when numerator equates to zero. That is:

$10\cdot x \cdot (x+3) - 5\cdot x^(2) = 0$

$10\cdot x^(2)+30\cdot x -5\cdot x^(2) = 0$

$5\cdot x^(2) + 30\cdot x = 0$

$5\cdot x \cdot (x+6) = 0$

This equation shows two critical points:

$x = 0$ , $x = -6$

c) The critical points found in point b) and the existence of a discontinuity in point a) lead to the conclusion of the existence local minima and maxima. By plotting the function, it is evident that $x = 0$ corresponds to a local maximum. (See Attachment)

d) By plotting the function, it is evident that $x = -6$ corresponds to a local minimum. (See Attachment)

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