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3 years ago

Consider the point.

Mathematics

1 answer:

aalyn [17]3 years ago

8 0

Answer:

L = 8.366

Step-by-step explanation:

Given

A(0,0,0)

H(3,5,6)

we can apply the equation

L = √((xH-xA)²+(yH-yA)²+(zH-zA)²)

where L is the length of the diagonal of the box

then

L = √((3-0)²+(5-0)²+(6-0)²)

⇒ L = √(9+25+36) = √70

⇒ L = 8.366

We can see the box in the pic shown.

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What is the scale factor in the dilation ? 1/6 1/3 3 6

anzhelika [568]

Answer:

Step-by-step explanation:

1) Find two corresponding lengths between the image and the pre-image

Pre-image (bottom side) = 3 units

Image (bottom side) = 9 units

We can use any corresponding lengths for this step.

2) Divide the length of the side from the image by the length of the side from the pre-image

9 units ÷ 3 units

= 3

Therefore, the scale factor of the dilation is 3.

I hope this helps!

5 0

2 years ago

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- A newspaper article discusses the rising prices of dairy products.it predicts that next year prices will be 115% of current pr

34kurt

Let x be the current price of the product

1.15 * x = $2..88

So when rearranging the equation, you get:
x = 2.88/1.15 = $2.50 (to the nearest cent)

8 0

2 years ago

Solve the system by graphing

julia-pushkina [17]

The graph shows the solution to be (x, y) = (-1, 3).

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3 years ago

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The graph of an equation is shown below:

olasank [31]

Answer:

(-2, -3)

Step-by-step explanation:

solutions are all the points where the line intersects

5 0

2 years ago

1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves

erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

$(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx$

$\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx$

$\ln|y| = x - \ln|x+1| + C$

When x = 0, we have y = 2, so we solve for the constant C :

$\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)$

Then the particular solution to the DE is

$\ln|y| = x - \ln|x+1| + \ln(2)$

We can go on to solve explicitly for y in terms of x :

$e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}$

(b) The curves y = x² and y = 2x - x² intersect for

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

and the bounded region is the set

$\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}$

The area of this region is

$\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}$

7 0

2 years ago