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3 years ago

Draw a model showing 15 divided by 3

Mathematics

1 answer:

kykrilka [37]3 years ago

6 0

Here is you're answer:

In order to solve this equation you must first solve the equation then make a model.

$15 \div 3 = 5$
$=5$

Draw three circles then divide 15 between the 3 circles to equal 5.

Therefore you're answer is "5."

Hope this helps!

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Solve the system by substitution.

Iteru [2.4K]

Answer:

I gave you most of the answer. I'll let you check my work and find the point using the solution.

Step-by-step explanation:

The first thing we do is we divide by negative one in the first equation to get

y = -x.

-3x + 3y = -36

Plug in y = -x and get

-3x + 3(-x) = -36

= -3x - 3x = -36

This equals -6x = -36

divide both sides by -6 and you get 6. 6 is your x value

Plug 6 back in to the second equation.

-3x + 3y = -36

-3(6) + 3y = -36

-18 + 3y = -36

3y = -18

y = -6

3 0

3 years ago

Jenn will use 18 connecting cubes to make a model of a park. The model will be in the shape of a rectangle and will have a heigh

Margarita [4]

This question is basically: how many different pairs of numbers multiply up to 18?

(This is a simpler form of the question because the question is just asking how many shapes of a rectangle can Jenn create with 18 blocks)

So, the numbers that multiply up to 18 are:

1 x 18
2 x 9
3 x 6

So, the answer is: Jenn can make the model of the park in 3 different ways.

3 0

4 years ago

What is the product of 14.23 and -5

wariber [46]

Answer: -71.15

Step-by-step explanation:

Multiply the two numbers

5 0

4 years ago

Read 2 more answers

Please someone help me with the question

Yuki888 [10]

Answer:

$\displaystyle \frac{d}{dx}[e^{2x}] = 2e^{2x}$

$\displaystyle \frac{d}{dx}[e^{3x}] = 3e^{3x}$

General Formulas and Concepts:

Algebra I

Terms/Coefficients
Exponential Rule [Multiplying]: $\displaystyle b^m \cdot b^n = b^{m + n}$

Calculus

Derivatives

Derivative Notation

eˣ Derivative: $\displaystyle \frac{d}{dx}[e^x] = e^x$

Derivative Rule [Product Rule]: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Step-by-step explanation:

Step 1: Define

$\displaystyle \frac{d}{dx}[e^{2x}] = \frac{d}{dx}[e^x \cdot e^x]$

$\displaystyle \frac{d}{dx}[e^{3x}] = \frac{d}{dx}[e^x \cdot e^{2x}]$

Step 2: Differentiate

$\displaystyle \frac{d}{dx}[e^{2x}]$

[Derivative] Product Rule: $\displaystyle \frac{d}{dx}[e^{2x}] = \frac{d}{dx}[e^x]e^x + e^x\frac{d}{dx}[e^x]$
[Derivative] eˣ Derivative: $\displaystyle \frac{d}{dx}[e^{2x}] = e^x \cdot e^x + e^x \cdot e^x$
[Derivative] Multiply [Exponential Rule - Multiplying]: $\displaystyle \frac{d}{dx}[e^{2x}] = e^{2x} + e^{2x}$
[Derivative] Combine like terms [Addition]: $\displaystyle \frac{d}{dx}[e^{2x}] = 2e^{2x}$

$\displaystyle \frac{d}{dx}[e^{3x}]$

[Derivative] Product Rule: $\displaystyle \frac{d}{dx}[e^{3x}] = \frac{d}{dx}[e^x]e^{2x} + e^x\frac{d}{dx}[e^{2x}]$
[Derivative] eˣ Derivatives: $\displaystyle \frac{d}{dx}[e^{3x}] = e^x(e^{2x}) + e^x(2e^{2x})$
[Derivative] Multiply [Exponential Rule - Multiplying]: $\displaystyle \frac{d}{dx}[e^{3x}] = e^{3x} + 2e^{3x}$
[Derivative] Combine like terms [Addition]: $\displaystyle \frac{d}{dx}[e^{3x}] = 3e^{3x}$

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Derivatives

Book: College Calculus 10e

8 0

3 years ago

2 questions please help me

Angelina_Jolie [31]

40) -31

41 ) 19

this are the answers I think

6 0

3 years ago