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3 years ago

To find 8 ×17, I double and halve

Mathematics

1 answer:

anastassius [24]3 years ago

6 0

Answer:

what is the question?

Step-by-step explanation:

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Valentin [98]

9514 1404 393

Answer:

A. -1 ≤ x ≤ 3

Step-by-step explanation:

The range is the horizontal extent of the graph--the set of x-values for which the function is defined. Here, the graph extends from x = -1 to x = 3, with both end points included. The range is ...

-1 ≤ x ≤ 3

5 0

3 years ago

The dimensions of a storage container in the shape of a rectangular prism are 54 in x 32.25 in x 24.5 in. Choose True or False f

AlladinOne [14]

GIVEN

the dimensions ofrectangular prism are

length = 54 in

breadth = 32.25in

height = 24.5 in

formula

volume of the rectangular prism = length×breadth×height

= 54 ×32.25×24.5

= 42997.5 in³

first statement is false because the volume of the rectangular prism in two significant digit.

Second statement is false because the volume of the rectangular prism in two significant digit.

FORMULA

surface area of a rectangular prism = 2(wl+hl+hw)

by putting the value of the length , breadth & height

we get

= 2 (54×32.25 + 32.25×24.5 + 24.5×54)

= 7709.25 in²

= 7709.3 in²(approx)

Third statement is true the surface area of the container is rounded to ones place.

Hence proved

3 0

4 years ago

Read 2 more answers

Use the substitution x = et to transform the given Cauchy-Euler equation to a differential equation with constant coefficients.

Anika [276]

Answer:

$\boxed{\sf \ \ \ ax^2+bx^{-10} \ \ \ }$

Step-by-step explanation:

Hello,

let's follow the advise and proceed with the substitution

first estimate y'(x) and y''(x) in function of y'(t), y''(t) and t

$x(t)=e^t\\\dfrac{dx}{dt}=e^t\\y'(t)=\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}=e^ty'(x)y'(x)=e^{-t}y'(t)\\y''(x)=\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}(e^{-t}\dfrac{dy}{dt})=-e^{-t}\dfrac{dt}{dx}\dfrac{dy}{dt}+e^{-t}\dfrac{d}{dx}(\dfrac{dy}{dt})\\=-e^{-t}e^{-t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}\dfrac{dt}{dx}=-e^{-2t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}e^{-t}\\=e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt})$

Now we can substitute in the equation

$x^2y''(x)+9xy'(x)-20y(x)=0\\ e^{2t}[ \ e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}) \ ] + 9e^t [ \ e^{-t}\dfrac{dy}{dt} \ ] -20y=0\\ \dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}+ 9\dfrac{dy}{dt}-20y=0\\ \dfrac{d^2y}{dt^2}+ 8\dfrac{dy}{dt}-20y=0\\$