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3 years ago

7

A 15 foot flagpole casts an 11 foot shadow. At the exact same time a 28 foot tree casts a shadow. Which proportion would correct

ly find the length of the tree's shadow?

1 answer:

coldgirl [10]3 years ago

8 0

15/11=28/x 15x=308 308÷15=20.5 3 reapeted

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rope price of length 45cm 25 cm and 81 cm have to be cut into same size pieces what is the smallest price length possible

Sergeeva-Olga [200]

= 2025

When you are told to find the smallest length possible, you perform L.C.M(Least common multiples)

For this, you divide the given lengths using the numbers that divides all through.

I have added an image to this answer. Go through it for more explanation

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2 years ago

Find the slope of the line passing through the points (-9, -6) and (-4, 5)

Masja [62]

Answer:

11/5 slope

Step-by-step explanation:

5-(-6) -4-(-9)

11/ 5

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2 years ago

Read 2 more answers

A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. 452 fee

Troyanec [42]

Answer:

school building, so the fourth side does not need Fencing. As shown below, one of the sides has length J.‘ (in meters}. Side along school building E (a) Find a function that gives the area A (I) of the playground {in square meters) in

terms or'x. 2 24(15): 320; - 2.x (b) What side length I gives the maximum area that the playground can have? Side length x : [1] meters (c) What is the maximum area that the playground can have? Maximum area: I: square meters

Step-by-step explanation:

8 0

2 years ago

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in <em>c)</em>

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution <em>m=α±βi</em> where <em>α=0</em> and <em>β=2</em>

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where <em>A</em> and <em>B</em> are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

7 0

3 years ago

X3 - 3x^2- 54x + 112

Olegator [25]

Answer:

(x -2) (x- 8) (x+7)

Step-by-step explanation:

6 0

2 years ago