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4 years ago

The diagram shows a right triangle and three squares. The area of the largest square is 55 units.

Mathematics

2 answers:

Keith_Richards [23]4 years ago

6 0

Answer:

Step-by-step explanation:

Well the sides of the smaller squares have to be greater than the biggest when put together.

a + b > c

So we need to find the square root of 55.

$\sqrt{55}$

≅ 7.4

Now for option A)

srt ⇒ 12 ≅3.5

srt ⇒ 43 ≅ 6.6

3.5 + 6.6 = 10.1

So option A is correct.

Option B)

srt ⇒ 14 ≅ 3.7

srt ⇒ 40 ≅ 6.3

3.7 + 6.3 = 10

So option B is also correct.

Option C)

srt ⇒ 16 = 4

srt ⇒ 37 ≅ 6.1

4 + 6.1 = 10.1

So option C is also correct.

Therefore,

all options are correct.

Hope this helps :)

ivann1987 [24]4 years ago

3 0

Answer:

It's 12 and 43

Step-by-step explanation:

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3 years ago

A rectangle has a perimeter of 30 ft. Find a function that models its area A in terms of the length x of one of its sides. What

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Answer:

Step-by-step explanation:

Let the other side of the rectangle be y. The perimeter of the rectangle is expressed as P = 2(x+y)

Given P = 30ft, on substituting P = 30 into the expression;

30 = 2(x+y)

x+y = 15

y = 15-x

Also since the area of the rectangle is xy;

A = xy

Substitute y = 15-x into the area;

A = x(15-x)

A = 15x-x²

The function that models its area A in terms of the length x of one of its sides is A = 15x-x²

The side of length x yields the greatest area when dA/dx = 0

dA/dx = 15-2x

15-2x = 0

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x = -15/-2

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Hence the side length, x that yields the greatest area is 7.5ft.

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3 years ago

I don't know how to do this

ValentinkaMS [17]

I would say c hope this helps

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3 years ago

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

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Answer:

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Step-by-step explanation:

Let's solve by separating variables:

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

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$dx=(t-sint)dt$

Apply integral both sides:

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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

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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 c)

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

This becomes the solution m=α±βi where α=0 and β=2

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

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Answer:

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