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2 years ago

I really need sum help plz

Mathematics

1 answer:

timama [110]2 years ago

3 0

Answer:

32 Strawberries

Step-by-step explanation:

The amount of strawberries you need for 48 cups of juice is 32

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V8 to the nearest tenth is about ?

777dan777 [17]

$\sqrt[]{8}\approx2.8$

6 0

1 year ago

What’s the distance between (4,-9) and (5,3)

laila [671]

Answer: Distance = √145

Concept:

Here, we need to know the concept of the distance formula.

The distance formula is the formula, which is used to find the distance between any two points.

If you are still confused, please refer to the attachment below for a clear version of the formula.

Solve:

Given information

(x₁, y₁) = (4, -9)

(x₂, y₂) = (5, 3)

Given formula

$Distance = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute values into the formula

$Distance = \sqrt{(5-4)^2+(3+9)^2}$

Simplify values in the parentheses

$Distance = \sqrt{(1)^2+(12)^2}$

Simplify exponents

$Distance = \sqrt{1+144}$

Simplify by addition

$Distance = \sqrt{145}$

Hope this helps!! :)

Please let me know if you have any questions

7 0

3 years ago

Read 2 more answers

A 1 newton force will stretch a spring 1 meter. The spring/mass system is damped by a force that is 8 times the instantaneous ve

sukhopar [10]

Answer:

Step-by-step explanation:

Given the mass is m =16kg, and 1N force will stretch the spring 1 m.

That is, F =1N,Z =1m. Now find the spring constant k:

F = kL = 1 = k(1) = k= 1N/m.

The damping force is 8times the instantaneous velocity, this means β = 8,

and the external force is f(t) = 0

Initially the object compressed 0.6m above equilibrium position,

with the downward velocity is 2m/s.

The differential equation for a spring mass system with

damping force and extemal force is: mx" + βxt + kx = f(t).

so, 16x"+ 8x' + x= 0, x(0} = -0.6, x'(0)= 2m/s.

Now solve the DE:

The auxilary equation for the homogeneous equation is 16x"+8x'+x=0

solving we get, 16r² + 8r + 1 = 0 => (4r + 1)² = 0 => r = - 1/4.

Then the general solution for the homogenous system is: $x(t)=c_1e^{-t/4} +c_2te^{-t\4}$ .

Use the initial conditions x (0) = -0.6, x'(0) = 2m/s:

$x(0)=c_1e^{0} +c_2(0)e^{0}=-0.6=c_1\\x'(t)=-\frac{1}{4}c_1e^{-t/4}+c_2e^{-t/4}-\frac{1}{4}c_2te^{-t/4}\\x'(0)=-\frac{1}{4}c_1e^0+c_2e^0-\frac{1}{4}c_2(0)e^0=2=-\frac{1}{4}(-0.6)+c_2=c_2=1.85$ .

Hence, $x(t) =-0.6e^{-t/4}+1.85te^{-t/4}$ .

5 0

3 years ago

Cards with the numbers 1 through 40 are placed in a box and thoroughly shuffled.

MrRa [10]

9/40 + 2/40 = 11/40

hope this helps

7 0

3 years ago

5x=y-4 write it in standard form

Tems11 [23]

5x - y = -4 is the standard form

7 0

3 years ago