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schepotkina [342]
2 years ago
15

Create at least two word problems that involve addition, subtraction, multiplication and division.

Mathematics
1 answer:
irinina [24]2 years ago
8 0

Answer:

One student  ate 21 sticks of gum in math class.  Each package of gum has 3 sticks of gum. How many packages of gum did he use?

21 /3 = 7 packages of gum

Step-by-step explanation:

John was having a contest to see how many pies he could eat in 5 minutes.

Every minute John can eat 12 pies.  So how many times did he eat in 5 minutes?

12 pies x 5 mins = 60 pies in the 5 minutes

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Is 4.95 a terminating decimal
Snowcat [4.5K]

Answer:

Definitely

since the definition of a terminating decimal is one that has an ending, and a Repeating decimal is one that never ends.

Step-by-step explanation:

6 0
3 years ago
Find the percent change.<br> 155 pounds to 130 pounds.
Romashka [77]

Answer:

130 pounds is about 84% of 155

8 0
3 years ago
What is 275.9 × 10−4 in standard form?
lisov135 [29]

Answer:

-2755

Step-by-step explanation:

4 0
4 years ago
1) Find the missing side. Give your<br> answer in simplest radical form.
gladu [14]

Answer:

x = 4√5

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

<u>Trigonometry</u>

[Right Triangles Only] Pythagorean Theorem: a² + b² = c²

  • a is a leg
  • b is another leg
  • c is the hypotenuse<u> </u>

Step-by-step explanation:

<u>Step 1: Define</u>

Leg <em>a</em> = 8

Leg <em>b</em> = 4

Hypotenuse <em>c</em> = <em>x</em>

<em />

<u>Step 2: Solve for </u><em><u>x</u></em>

  1. Substitute in variables [Pythagorean Theorem]:                                            8² + 4² = x²
  2. Evaluate exponents:                                                                                         64 + 16 = x²
  3. Add:                                                                                                                   80 = x²
  4. [Equality Property] Square root both sides:                                                    √80 = x
  5. Rewrite:                                                                                                             x = √80
  6. Simplify:                                                                                                             x = 4√5
3 0
3 years ago
) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra
artcher [175]

Answer:

y(t)=2e^{3t}(2-5t)

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2  

Using the theorem of the Laplace transform for derivatives, we know that:

\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)

Replacing the initial values y(0)=4, y′(0)=2 we obtain

\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4

and our differential equation (*) gets transformed in the algebraic equation

\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0

Solving for Y(s) we get

\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}

Now, we brake down the rational expression of Y(s) into partial fractions

\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here  

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}

and

\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}

we know that

\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}

and for the first translation property of the inverse Laplace transform

\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}

and the solution of our differential equation is

\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}

5 0
3 years ago
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