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3 years ago

25. Find the length of a rectangle with a perimeter of 14 and a width of 10

Mathematics

1 answer:

Genrish500 [490]3 years ago

6 0

Answer:

2 (added together it's 4)

Step-by-step explanation:

Assuming the width of 10 was both sides added together that makes it 5+5

14-10=4

4/2=2

so the sides should be 5,2,5,2

the wording is the most difficult thing about this question.

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For an outdoor concert by the city orchestra, concert organizers estimate that 16 comma 000 people will attend if it is not rai

Sladkaya [172]

Answer: 13300

========================================

Work Shown:

A = event that it rains

B = event that it does not rain

P(A) = 0.30

P(B) = 1-P(A) = 1-0.30 = 0.70

Multiply the attendance figures with their corresponding probabilities

if it rains, then 7000*P(A) = 7000*0.30 = 2100
if it doesn't rain, then 16000*P(B) = 16000*0.70 = 11200

Add up the results: 2100+11200 = 13300

This is the expected value. This is basically the average based on the probabilities. The average is more tilted toward the higher end of the spectrum (closer to 16000 than it is to 7000) because there is a higher chance that it does not rain.

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3 years ago

What is h(-2)<br> H(-1.75)<br> H(1)

ruslelena [56]

$h(x)=\left\{\begin{array}{cccc}-2&if&-2\leq x$

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4 years ago

A bottle of fruit juice holds 1.89L. If Shakira bought almost 6L of fruit juice, how many bottles did she buy

ser-zykov [4K]

6 divided by 1.89= 3.17 so you can say she bought 3 bottles of fruit juice

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3 years ago

find the coordinates of the point P on the parabola y=1-x^2 with domain 0≤x≤1 that minimize the area of the triangle enclosed by

coldgirl [10]

Let point P be with coordinates $(x_0,y_0).$ Find the equation of the tangent line.

1. If $y=1-x^2,$ then $y'=-2x.$

2. The equation of the tangent line at point P is

$y-y_0=-2x_0(x-x_0).$

Find x-intercept and y-intercept of this line:

when x=0, then $y=y_0+2x_0^2;$
when y=0, then $x=\dfrac{y_0}{2x_0}+x_0=\dfrac{y_0+2x_0^2}{2x_0}.$

The area of the triangle enclosed by the tangent line at P, the x-axis, and y-axis is

$A=\dfrac{1}{2}\cdot (2x_0^2+y_0)\cdot \left(\dfrac{y_0+2x_0^2}{2x_0}\right)=\dfrac{(y_0+2x_0^2)^2}{4x_0}.$

Since point P is on the parabola, then $y_0=1-x_0^2$ and

$A=\dfrac{(1-x_0^2+2x_0^2)^2}{4x_0}=\dfrac{(1+x_0^2)^2}{4x_0}.$

Find the derivative A':

$A'=\dfrac{2(1+x_0^2)\cdot 2x_0\cdot 4x_0-4(1+x_0^2)^2}{16x_0^2}=\dfrac{12x_0^4+8x_0^2-4}{16x_0^2}.$

Equate this derivative to 0, then

$12x_0^4+8x_0^2-4=0,\\ \\3x_0^4+2x_0^2-1=0,\\ \\D=2^2-4\cdot 3\cdot (-1)=16,\ \sqrt{D}=4,\\ \\x_0^2_{1,2}=\dfrac{-2\pm4}{6}=-1,\dfrac{1}{3},\\ \\x_0^2=\dfrac{1}{3}\Rightarrow x_0_{1,2}=\pm\dfrac{1}{\sqrt{3}}.$