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Anvisha [2.4K]
3 years ago
7

Arrange the equations in the correct sequence to rewrite the formula for displacement, , to find a. In this formula, d is displa

cement, v0 is initial velocity, a is acceleration, and t is time.
Mathematics
1 answer:
valina [46]3 years ago
8 0
Answer: 

a=2[d-V_0t]/t^2=2d/t^2-2v_0/t

Explanation:


Since, you have not included the formula, I will work here with the formula for constant accelaration motion that relates the four variables: displacement (d), Vo (initial velocity), a (acceleration) and t (time).

1) displacement formula:

d= V_0t+\frac{1}{2} at^2

2) Subtract the term Vot from both sides:

d-V_0t= \frac{1}{2} at^2

3) Multiply both sides by 2:

2d -2V_0t=at^2

4) Divide both sides by t²

2[d-V_0t]/t^2=a

So, you have obtainded:

a = 2[d - Vo×t] / t²

Yet, you can arrange it in different ways. For example, you might separate into two terms:

a =  \frac{2d}{t^2} - \frac{2V_0}{t}

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What is the answer to 4x-5x+9=5x-9?
fiasKO [112]

Answer:

x=3

Step-by-step explanation:

First, combine like terms on the left side

4x-5x+9=5x-9

-x+9=5x-9

Subtract 9 on both sides

-x=5x-18

Subtract 5x on both sides

-6x= -18

Divide by -6 on both sides

x=3

Hope this helps! :)

5 0
4 years ago
Read 2 more answers
Help plzzz<br>Select the correct answer.<br>​
Morgarella [4.7K]
<h3>Answer:  C) 3</h3>

=========================================================

Explanation:

f(x) is the outer function, so the final output -8 corresponds to f(x)

We see that f(-4) = -8 in the first column of the table. I'm starting with the output and working my way backward to get the input. So we started with -8 and worked back to -4.

Then we move to the g(x) function to follow the same pattern: start with the output and move to the input. We start at -4 in the g(x) bubble and move to 3 in the x bubble.

In short, g(3) = -4

So,

f(g(x)) = f(g(3)) = f(-4) = -8

We see that x = 3 leads to f(g(x)) = -8

3 0
3 years ago
[(7+3)x5-4](divided by) 2+3<br> SHOW ALL WORK WILL MARK BRAINLIEST
Nesterboy [21]

Answer:

26

Step-by-step explanation:

[(7+3)5-4]/2+3

-To solve this equation you have to use PEMDAS

P- Parentheses

E- Exponents

M- Multiplication

D- Division

A- Addition

S- Subtraction-

- With MD and AS you work left to right of the equation since they are in the same spot. (PE[MD][AS])

Step 1) [(10)5-4]/2+3

- First you do "P," parentheses, so you add 7+3=10

Step 2) [50-4]/2+3

- Next you do "M," multiplication, and multiply 10x5=50

Step 3) [46]/2+3

- Then you do "S," subtraction, and subtract 50-4=46

(FYI: Steps 1-3 were still in the parentheses. We had to start with the parentheses in the parentheses, work PEMDAS, and now we are out of the parentheses and have to work PEMDAS on the rest of the problem.)

Step 4) 23+3

- Now we do "D," division, and divide 46/2=23

Step 5) 23+3=6

- Finally we do "A," addition, and add 23+3=26 so the answer is 26

(FYI: "/" means division)

4 0
3 years ago
Consider the function f given by f(x)=x*(e^(-x^2)) for all real numbers x.
NISA [10]

Answer:

\frac{\sqrt{\pi}}{4}

Step-by-step explanation:

You are going to integrate the following function:

g(x)=x*f(x)=x*xe^{-x^2}=x^2e^{-x^2}  (1)

furthermore, you know that:

\int_0^{\infty}e^{-x^2}=\frac{\sqrt{\pi}}{2}

lets call to this integral, the integral Io.

for a general form of I you have In:

I_n=\int_0^{\infty}x^ne^{-ax^2}dx

furthermore you use the fact that:

I_n=-\frac{\partial I_{n-2}}{\partial a}

by using this last expression in an iterative way you obtain the following:

\int_0^{\infty}x^{2s}e^{-ax^2}dx=\frac{(2s-1)!!}{2^{s+1}a^s}\sqrt{\frac{\pi}{a}} (2)

with n=2s a even number

for s=1 you have n=2, that is, the function g(x). By using the equation (2) (with a = 1) you finally obtain:

\int_0^{\infty}x^2e^{-x^2}dx=\frac{(2(1)-1)!}{2^{1+1}(1^1)}\sqrt{\pi}=\frac{\sqrt{\pi}}{4}

5 0
3 years ago
Read 2 more answers
Solve 6/×-3 =x/3
Eduardwww [97]

Answer:

X = -3. non-extraneous

Step-by-step explanation:

The given equation is \frac{6}{x-3} =\frac{x}{3}

We cross multiply  to get:

6*3=x(x-3)

Expand and multiply.

18=x^2-3x

x^2-3x-18=0

We factor to obtain:

(x-6)(x+3)=0

x=-3,x=6

Checking for extraneous solutions

Substitute x=-3 \frac{6}{-3-3} =\frac{-3}{3}

\frac{6}{-6} =\frac{-3}{3}

-1=-1

1=1.....Non-extraneous

Checking for x=6

\frac{6}{6-3} =\frac{6}{3}

\frac{6}{3} =\frac{6}{3}

2=2....Non-extraneous

6 0
3 years ago
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