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Vitek1552 [10]
3 years ago
5

Can sum1 help me with this question

Mathematics
2 answers:
lutik1710 [3]3 years ago
8 0

Answer:

Second option is the correct choice.

Step-by-step explanation:

Thanks!

larisa86 [58]3 years ago
6 0
Height of the balls after two bounces is

52/100 × (0.52 ×0.5) = 0.52 × 0.52 × 0.5
= 0.52^2 × 0.5 m

Therefore the series turns into 0.5,0.5(0.52),0.5(0.52)^2,....

So by finding the height f(n) after n bounces would be
f(n) = 0.5(0.52)^n-1

To get the height of the ball of the third path, we need to put n=3

f(3)= 0.5(0.52)^3-1
=0.5(0.52)^2
=0.14m

Answer: B ) 0.5 × (0.52)^n-1 ; 0.14m
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A second-order linear differential equation for y(t) is said to be homogeneous if... every term involve either y or its derivati
swat32

Answer: Hello!

A second order differential equation has the next shape:

y''(t) + p(t)y'(t) + q(t)y(t) = g(t)

where p(t), q(t) and g(t) are functions of t, that can be constant numbers for example.

And is called homogeneus when g(t) = 0, so you have:

y''(t) + p(t)y'(t) + q(t)y(t) = 0

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3 years ago
Carl is going to make a frame for a picture. The dimensions of the frame need to be 2 feet by 18 inches. How long must the board
MArishka [77]
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3 years ago
What's the value of c that makes 9x^2 + 12x + c a perfect square trinomial.
seraphim [82]
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4 0
3 years ago
How to find imaginary zeros anf real zeros of F(x)=-4x^5-8x^3+12x​
Fofino [41]

Answer:

x=\{0, -1, 1, -i\sqrt{3}, i\sqrt{3}\}

Step-by-step explanation:

We are given the function:

f(x)=-4x^5-8x^3+12x

And we want to finds its zeros.

Therefore:

0=-4x^5-8x^3+12x

Firstly, we can divide everything by -4:

0=x^5+2x^3-3x

Factor out an x:

0=x(x^4+2x^2-3)

This is in quadratic form. For simplicity, we can let:

u=x^2

Then by substitution:

0=x(u^2+2u-3)

Factor:

0=x(u+3)(u-1)

Substitute back:

0=x(x^2+3)(x^2-1)

By the Zero Product Property:

x=0\text{ and } x^2+3=0\text{ and } x^2-1=0

Solving for each case:

x=0\text{ and } x=\pm\sqrt{-3}\text{ and } x=\pm\sqrt{1}

Therefore, our real and complex zeros are:

x=\{0, -1, 1, -i\sqrt{3}, i\sqrt{3}\}

4 0
2 years ago
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