Make a table and graph y=x2-4

The sides of two cubes are in the ratio 2:1. What is the ratio of their volumes ?

olasank [31]

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Answer:

we know that, Volume of cube=a³

Let volume of 1st cube be 2x with side a and other be x with side A (according to given ratio)

Step-by-step explanation:

ATQ: 2x=a³ and, x=A³

a=³√2x and, A=x

we know that, surface area of cube is 6a²

Thus, surface area of 1st cube = 6(³√2x)²

= 6³√4x²

Surface area of 2nd cube=6(x)²=6x²

Ratio of S.A=(6³√4x²)÷(6x²)

Ratio of S.A=³√4:1

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5 0

3 years ago

F(x) = {1 if x < 1

son4ous [18]

Have a wonderful day today

7 0

4 years ago

1. The number of points scored in a game varies directly to the number of minutes played. After

rusak2 [61]

Answer:

Step-by-step explanation:

1a:

3/1. 24/8 simplifies to 3/1.

1b:

18 points. The numerator and denominator of 3/1 can be multiplied by 18 to get 54/18.

2a:

2.45/1. 73.3/30 simplifies to 2.45/1.

2b: I don't really know, someone else might have to answer, but I hope what I did put helps!

5 0

3 years ago

.What decimal is being represented by the decimal grid?

Lisa [10]

It’s 2.26 (now I am going to write random stuff bc it says I need to have 20 characters ggvjffdffvhkjhv)

4 0

3 years ago

If w = 5 cos (xy) − sin (xz) and x = 1/t , y = t, z = t^3 ; then find dw/dt

Scrat [10]

In this question, we find the derivatives, using the chain's rule.

Doing this, the derivative is:

$\frac{dw}{dt} = \frac{5}{t}(\sin{1} - \cos{1}) - 2t\cos{t^2}$

Chain Rule:

Suppose we have a function $w(x,y,z), x = x(t), y = y(t), z = z(t)$ , and want to find it's derivative as function of t. It will be given by:

$\frac{dw}{dt} = \frac{dw}{dx}\frac{dx}{dt} + \frac{dw}{dy}\frac{dy}{dt} + \frac{dw}{dz}\frac{dz}{dt}$

Thus, we have to find the desired derivatives, which are:

w of x:

$\frac{dw}{dx} = -5y\sin{(xy)} - z\cos{(xz)}$

Considering $x = \frac{1}{t}, y = t, z = t^3$

$\frac{dw}{dx} = -5t\sin{(1)} - t^3\cos{(t^2)}$

w of y:

$\frac{dw}{dy} = -5x\cos{(xy)}$

Considering $x = \frac{1}{t}, y = t$

$\frac{dw}{dy} = -\frac{5}{t}\cos{1}$

w of z:

$\frac{dw}{dz} = -x\cos{(xz)}$

Considering $x = \frac{1}{t}, z = t^3$

$\frac{dw}{dz} = -\frac{1}{t}\cos{(t^2)}$

Derivatives of x, y and z as functions of t:

$\frac{dx}{dt} = -\frac{1}{t^2}$

$\frac{dy}{dt} = 1$

$\frac{dz}{dt} = 3t^2$

Derivative of w as function of t.

Now, we just replace what we found into the formula. So

$\frac{dw}{dt} = \frac{dw}{dx}\frac{dx}{dt} + \frac{dw}{dy}\frac{dy}{dt} + \frac{dw}{dz}\frac{dz}{dt}$

$\frac{dw}{dt} = (-5t\sin{(1)} - t^3\cos{(t^2)})(-\frac{1}{t^2}) - (\frac{5}{t}\cos{1}) - (\frac{1}{t}\cos{(t^2)})3t^2$

Applying the multiplications:

$\frac{dw}{dt} = \frac{5}{t}\sin{1} + t\cos{t^2} - \frac{5}{t}\cos{1} - 3t\cos{t^2}$

Applying the simplifications:

$\frac{dw}{dt} = \frac{5}{t}(\sin{1} - \cos{1}) - 2t\cos{t^2}$

Which is the derivative.

For more on the chain rule, you can check brainly.com/question/12795383

8 0

3 years ago