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3 years ago

What is the largest perimeter possible for a rectangle with positive whole number dimensions and an area of 40 cm?

Mathematics

1 answer:

Vladimir79 [104]3 years ago

8 0

Answer:

C, 82 cm

Step-by-step explanation:

If the area of a rectangle is it's length times it's width and for the problem, it's maximum area must be 40cm than 40x1 would offer the largest perimeter with the smallest area respectively totalling at 82 cm.

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3. The curve C with equation y=f(x) is such that, dy/dx = 3x^2 + 4x +k

Andreas93 [3]

a. Given that y = f(x) and f(0) = -2, by the fundamental theorem of calculus we have

$\displaystyle \frac{dy}{dx} = 3x^2 + 4x + k \implies y = f(0) + \int_0^x (3t^2+4t+k) \, dt$

Evaluate the integral to solve for y :

$\displaystyle y = -2 + \int_0^x (3t^2+4t+k) \, dt$

$\displaystyle y = -2 + (t^3+2t^2+kt)\bigg|_0^x$

$\displaystyle y = x^3+2x^2+kx - 2$

Use the other known value, f(2) = 18, to solve for k :

$18 = 2^3 + 2\times2^2+2k - 2 \implies \boxed{k = 2}$

Then the curve C has equation

$\boxed{y = x^3 + 2x^2 + 2x - 2}$

b. Any tangent to the curve C at a point (a, f(a)) has slope equal to the derivative of y at that point:

$\dfrac{dy}{dx}\bigg|_{x=a} = 3a^2 + 4a + 2$

The slope of the given tangent line $y=x-2$ is 1. Solve for a :

$3a^2 + 4a + 2 = 1 \implies 3a^2 + 4a + 1 = (3a+1)(a+1)=0 \implies a = -\dfrac13 \text{ or }a = -1$

so we know there exists a tangent to C with slope 1. When x = -1/3, we have y = f(-1/3) = -67/27; when x = -1, we have y = f(-1) = -3. This means the tangent line must meet C at either (-1/3, -67/27) or (-1, -3).

Decide which of these points is correct:

$x - 2 = x^3 + 2x^2 + 2x - 2 \implies x^3 + 2x^2 + x = x(x+1)^2=0 \implies x=0 \text{ or } x = -1$

So, the point of contact between the tangent line and C is (-1, -3).

7 0

2 years ago

Please help with this simple trig problem- angle of two sectors of a circle

igomit [66]

Answer:

K = (1/2)r^2(sin(θ) +θ)

Step-by-step explanation:

The area of the triangle to the left is ...

A1 = (1/2)r^2·sin(180°-θ) = (1/2)r^2·sin(θ)

The area of the sector to the right is ...

A2 = (1/2)r^2θ

so the total area of the blue shaded region is ...

K = A1 + A2 = (1/2)r^2·sin(θ) + (1/2)r^2·θ

K = (1/2)r^2(sin(θ) +θ)

4 0

3 years ago

What is the slope of the line?

Lelu [443]

Answer:

The slope is 2. (m=2).

Step-by-step explanation:

y+5=2x+2

y+5-5=2x+2-5

y=2x-3

Equation is y=2x-3.

y=y-intercept

x=slope

Hence, the slope is 2.

Hope this helps and stay safe!

8 0

3 years ago

A 20 kg bag of bananas for $10<br><br><br> ______ per kg

NeTakaya

Answer:

each bannana i think

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

What is the midpoint of the segment below?

ZanzabumX [31]

Answer:

Step-by-step explanation:

(5+(-4))/2 = 1/2 or 0.5

(-7 + 6)/2 = -1/2 or -0.5

the solution is D

(0.5, -0.5)

5 0

3 years ago