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4 years ago

5

How do we transform system a into system b? THIS IS KOT MULTIPLE CHOICE. PLEASE ANSWER ASAP!!

1 answer:

Alina [70]4 years ago

3 0

JESUS IS ALWAYS THE ANSWER

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Find the area of the regular 15-gon with radius 10mm.

Nady [450]

318.83484250503 mm^2

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3 years ago

Finding angle measures between intersecting lines

Doss [256]

Answer:

x = 60°

Step-by-step explanation:

G= 90°

30 + 90 = 120

180 - 120 = 60

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3 years ago

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assume x and y are both differentiable functions of t. find dx/dt given x=-1 and dy/dt=8 for the relation: 4x^2+3y^3=28

melomori [17]

Given:

x and y are both differentiable functions of t.

$4x^2+3y^3=28$

$x=-1\text{ and }\dfrac{dy}{dt}=8$

To find:

The value of $\dfrac{dx}{dt}$ .

Solution:

We have,

$4x^2+3y^3=28$ ...(i)

At x=-1,

$4(-1)^2+3y^3=28$

$4+3y^3=28$

$3y^3=28-4$

$3y^3=24$

Divide both sides by 3.

$y^3=8$

Taking cube root on both sides.

$y=2$

So, y=2 at x=-1.

Differentiate (i) with respect to t.

$4(2x\dfrac{dx}{dt})+3(3y^2\dfrac{dy}{dt})=0$

Putting x=-1, y=2 and $\dfrac{dy}{dt}=8$ , we get

$4(2(-1)\dfrac{dx}{dt})+3(3(2)^28)=0$

$-8\dfrac{dx}{dt}+9(4)(8)=0$

$-8(\dfrac{dx}{dt}-9(4))=0$

Divide both sides by -8.

$\dfrac{dx}{dt}-36=0$

$\dfrac{dx}{dt}=36$

Therefore, the value of $4x^2+3y^3=28$ is 36.

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3 years ago

Factor 1 - 2.25x 8.......

garri49 [273]

1-2.25x8=(1-1.5x^4)(1+1.5x^4)
=(1-√1.5 x²)(1+√1.5 x²)(1+1.5 x^4)
=(1-√√1.5 x)(1+√√1.5 x)(1+√1.5 x²) (1+√1.5 x^4)

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4 years ago

Read 2 more answers

3) Postcards: Marcie wants to make a box to hold her postcard collection from a piece of

klasskru [66]

The volume of the box is the amount of space in it.

The dimension of the postcard is:

$\mathbf{Length = 10}$

$\mathbf{Width = 18}$

Let x represents the length of the cut.

So, the dimension of the box is:

$\mathbf{Length = 10 -2x}$

$\mathbf{Width = 18 -2x}$

$\mathbf{Height = x}$

<u>(a) The function that represents the volume of the box</u>

This is calculated as:

$\mathbf{V(x) = Length \times Width \times Height}$

So, we have:

$\mathbf{V(x) = (10 - 2x) \times (18 - 2x) \times x}$

Expand

$\mathbf{V(x) = 180x - 56x^2 + 4x^3}$

Hence, the volume function is: $\mathbf{V(x) = 180x - 56x^2 + 4x^3}$

<u>(b) The dimension that maximizes the volume</u>

We have:

$\mathbf{V(x) = 180x - 56x^2 + 4x^3}$

Differentiate

$\mathbf{V'(x) = 180- 112x + 12x^2}$

Set to 0

$\mathbf{180- 112x + 12x^2 = 0}$

Using a calculator, we have:

$\mathbf{x = 7.3,2.1}$

The value x = 7.3 is greater than the dimension of the box.

So, we have:

$\mathbf{x = 2.1}$

Recall that:

$\mathbf{Length = 10 -2x}$

$\mathbf{Width = 18 -2x}$

$\mathbf{Height = x}$

So, we have:

$\mathbf{Length = 10 -2 \times 2.1 = 5.8}$

$\mathbf{Width = 18 -2 \times 2.1 = 13.8}$

$\mathbf{Height = 2.1}$

Hence, the dimension that maximizes the volume is 5.8 by 13.8 by 2.1

<u>(c) The maximum volume</u>

This is calculated as the product of the dimension of the box

So, we have:

$\mathbf{Volume = 5.8 \times 13.8 \times 2.1}$

$\mathbf{Volume = 168.1}$

Hence, the maximum volume of the box is 168.1 cubic inches

Read more about volumes at:

brainly.com/question/17056191

5 0

3 years ago