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3 years ago

Which of these is an example of physical change?

1 answer:

8 0

a digesting bread because you're not changing anything about the you're just eating the bread and eating the bread has no change

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A water pipe is 4 7/12 feet beneath the surface of the road. What is the elevation of the pipe, in feet, expressed as a decimal?

Andrej [43]

The answer is C. U just have to divide 7 and 12 to get your decimal

8 0

3 years ago

Help me with this please ASAP

Natasha2012 [34]

Answer:

Hi there your answer is $8x^{6}$

Step-by-step explanation:

Hi... You have to muliply 4 and 2 which will equal to 8

Then, if you know the exponent rule for the Product Rule or don't I'll still give it you here it is

$x^{m}*x^{n}=x^{m+n}$

In this case we will be adding 3 and 3 which will equal to 6

Which your answer will be

$8x^{6}$

4 0

3 years ago

Expert Answer Please !Decide If SSS,ASA,AAS,SAS,HL, or None

ValentinkaMS [17]

Sss , ik this because all u have to do is read the question , self explanatory

3 0

3 years ago

Which fractions are equivalent to 4/6? Select all that apply.

kiruha [24]

Answer:

2/3

8/12

Step-by-step explanation:

You just divide each numerator 4 (5÷4) and each denominator by 6 (8÷6) and see if the numbers are the same.

5÷4=1.25

8÷6=1.33

They are not equivalent, so 5/8 is not correct

2÷4=0.5

3÷6=0.5

They are equivalent, so 2/3 is correct

Please give brainliest if I helped! :)

4 0

3 years ago

Solve the initial-value problem using the method of undetermined coefficients.

andrey2020 [161]

First check the characteristic solution. The characteristic equation to this DE is

r ² - r = r (r - 1) = 0

with roots r = 0 and r = 1, so the characteristic solution is

y (char.) = C₁ exp(0x) + C₂ exp(1x)

y (char.) = C₁ + C₂ exp(x)

For the particular solution, we try the ansatz

y (part.) = (ax + b) exp(x)

but exp(x) is already accounted for in the second term of y (char.), so we multiply each term here by x :

y (part.) = (ax ² + bx) exp(x)

Differentiate this twice and substitute the derivatives into the DE.

y' (part.) = (2ax + b) exp(x) + (ax ² + bx) exp(x)

… = (ax ² + (2a + b)x + b) exp(x)

y'' (part.) = (2ax + 2a + b) exp(x) + (ax ² + (2a + b)x + b) exp(x)

… = (ax ² + (4a + b)x + 2a + 2b) exp(x)

(ax ² + (4a + b)x + 2a + 2b) exp(x) - (ax ² + (2a + b)x + b) exp(x)

= x exp(x)

The factor of exp(x) on both sides is never zero, so we can cancel them:

(ax ² + (4a + b)x + 2a + 2b) - (ax ² + (2a + b)x + b) = x

Collect all the terms on the left side to reduce it to

2ax + 2a + b = x

Matching coefficients gives the system

2a = 1

2a + b = 0

and solving this yields

a = 1/2, b = -1

Then the general solution to this DE is

y(x) = C₁ + C₂ exp(x) + (1/2 x ² - x) exp(x)

For the given initial conditions, we have

y (0) = C₁ + C₂ = 6

y' (0) = C₂ - 1 = 5

and solving for the constants here gives

C₁ = 0, C₂ = 6

so that the particular solution to the IVP is

y(x) = 6 exp(x) + (1/2 x ² - x) exp(x)

3 0

3 years ago