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3 years ago

What is my weighted grade please explain how it's done.

1 answer:

4 0

That's not enough information to calculate your grade from.

-- You've told us the scores you have in things that go together
to make up (35 + 5 + 25) = 65% of your grade.

-- We don't know your scores in the things that make up the other
35% of your grade.

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Two numbers are of the ratio 3:5. If their difference is 18, find the numbers.

Sergio039 [100]

$x \div y = 3 \div 5 \\ it \: means = \\ 3y = 5x \\ x - y = 18 \\$

To solve we multiply the last equation by 5 and we get the following:

5x-5y=90

But 5x=3y and we replace it in this equation:

3y-5y=90

-2y=90

y=-45

and x-(-45)=18 , x+45=18

x=-27

4 0

3 years ago

If 1/2 gallon of paint covers 1/4 of a wall then how much paint is needed to cover the entire wall

AleksAgata [21]

Answer:

Step-by-step explanation:

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=3%20-%20%20%5Cfrac%7Bx%7D%7B2%7D%20%20%3C%203%20" id="TexFormula1" title="3 - \frac{x}{2} &l

Sergio [31]

The first step for solving this is to cancel equal terms on both sides of the inequality.
$- \frac{x}{2}$ < 0
Now multiply both sides of the inequality by -2 and flip the inequality sign.
-2 × ( $- \frac{x}{2}$ ) > -2 × 0
Remember that multiplying two negatives together always equals a positive,, so the expression changes to the following:
2 × $\frac{x}{2}$ > -2 × 0
Reduce the numbers with 2.
x > -2 × 0
Any expression multiplied by 0 equals 0,, so the correct answer to your question will be:
x > 0
Let me know if you have any further questions.
:)

8 0

3 years ago

In this problem you will use undetermined coefficients to solve the nonhomogeneous equation y′′+4y′+3y=8te^−t+6e^−t−(9t+6)

Luden [163]

We're given the ODE,

y'' + 4y' + 3y = 8t exp(-t ) + 6 exp(-t ) - (9t + 6)

(where I denote exp(x) = eˣ )

First determine the characteristic solution:

y'' + 4y' + 3y = 0

has characteristic equation

r ² + 4r + 3 = (r + 1) (r + 3) = 0

with roots at r = -1 and r = -3, so the characteristic solution is

y = C₁ exp(-t ) + C₂ exp(-3t )

For the non-homogeneous equation, assume two ansatz solutions

y₁ = (at ² + bt + c) exp(-t )

and

y₂ = at + b

• y'' + 4y' + 3y = 8t exp(-t ) + 6 exp(-t ) … … … [1]

Compute the derivatives of y₁ :

y₁ = (at ² + bt + c) exp(-t )

y₁' = (2at + b) exp(-t ) - (at ² + bt + c) exp(-t )

… = (-at ² + (2a - b) t + b - c) exp(-t )

y₁'' = (-2at + 2a - b) exp(-t ) - (-at ² + (2a - b) t + b - c) exp(-t )

… = (at ² + (b - 4a) t + 2a - 2b + c) exp(-t )

Substitute them into the ODE [1] to get

→ [(at ² + (b - 4a) t + 2a - 2b + c) + 4 (-at ² + (2a - b) t + b - c) + 3 (at ² + bt + c)] exp(-t ) = 8t exp(-t ) + 6 exp(-t )

(at ² + (b - 4a) t + 2a - 2b + c) + 4 (-at ² + (2a - b) t + b - c) + 3 (at ² + bt + c) = 8t + 6

4at + 2a + 2b = 8t + 6

→ 4a = 8 and 2a + 2b = 6

→ a = 2 and b = 1

→ y₁ = (2t ² + t ) exp(-t )

(Note that we don't find out anything about c, but that's okay since it would have gotten absorbed into the first characteristic solution exp(-t ) anyway.)

• y'' + 4y' + 3y = -(9t + 6) … … … [2]

Compute the derivatives of y₂ :

y₂ = at + b

y₂' = a

y₂'' = 0

Substitute these into [2] :

4a + 3 (at + b) = -9t - 6

3at + 4a + 3b = -9t - 6

→ 3a = -9 and 4a + 3b = -6

→ a = -3 and b = 2

→ y₂ = -3t + 2

Then the general solution to the original ODE is

y(t) = C₁ exp(-t ) + C₂ exp(-3t ) + (2t ² + t ) exp(-t ) - 3t + 2

Use the initial conditions y (0) = 2 and y' (0) = 2 to solve for C₁ and C₂ :

y (0) = C₁ + C₂ + 2 = 2

→ C₁ + C₂ = 0 … … … [3]

y'(t) = -C₁ exp(-t ) - 3C₂ exp(-3t ) + (-2t ² + 3t + 1) exp(-t ) - 3

y' (0) = -C₁ - 3C₂ + 1 - 3 = 2

→ C₁ + 3C₂ = -4 … … … [4]

Solve equations [3] and [4] to get C₁ = 2 and C₂ = -2. Then the particular solution to the initial value problem is

y(t) = -2 exp(-3t ) + (2t ² + t + 2) exp(-t ) - 3t + 2

7 0

2 years ago

Write 2.9687 in p/q form , where p and q are integers and q is non zero

Anna35 [415]

Answer:

Step-by-step explanation:

2.9687= $\frac{29687}{10000}$

4 0

3 years ago