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REY [17]
3 years ago
15

What is the density of a material that had a mass of 2500 grams and a volume of 500 cubic centimeters,in grams per cubic centime

ter?
Mathematics
1 answer:
lorasvet [3.4K]3 years ago
6 0
2500/500=5

2500 grams/500 cm^3 = 5 g/cm^3
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3x^2-x=10<br><br> Find factors <br><br><br> Determine zeros
Sophie [7]

Factors are (3x + 5) and (x - 2). Zeros are -5/3 and 2.

Step-by-step explanation:

  • Step 1: Given quadratic equation 3x² - x = 10 ⇒ 3x² - x - 10 = 0
  • Step 2: Use factoring method (product and sum rule) to find factors and zeros. So the equation can be written as below.

⇒ 3x² - 6x + 5x - 10 = 0

⇒ 3x(x - 2) + 5(x - 2) = 0

⇒ (3x + 5)(x - 2) = 0 These are the factors of the equation.

  • Step 3: Find zeros by equating the factors to 0.

⇒ 3x + 5 = 0 and x - 2 = 0

⇒ x = -5/3 and 2

7 0
3 years ago
Given j(x) = x + 5, what is the value of j(12)?
yarga [219]

Answer:

j(12) = 17

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

<u>Algebra I</u>

  • Function Notation

Step-by-step explanation:

<u>Step 1: Define</u>

j(x) = x + 5

j(12) is x = 12

<u>Step 2: Evaluate</u>

  1. Substitute in <em>x</em>:                    j(12) = 12 + 5
  2. Add:                                     j(12) = 17
8 0
3 years ago
Read 2 more answers
What is <img src="https://tex.z-dn.net/?f=3%5E%7Bx%7D%20%3D%20%5Cfrac%7B1%7D%7B9%7D" id="TexFormula1" title="3^{x} = \frac{1}{9}
Makovka662 [10]

~\hspace{7em}\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} ~\hspace{4.5em} a^n\implies \cfrac{1}{a^{-n}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 3^x=\cfrac{1}{9}\implies 3^x=\cfrac{1}{3^2}\implies 3^x=3^{-2}\implies x = -2

8 0
2 years ago
A purchase of supplies for cash is recorded in the
ANEK [815]

Answer:

cash payment journal

Step-by-step explanation:

3 0
3 years ago
Solving systems elimination: I don’t understand how to solve at all. Please help if you can.
german

The solution to given system of equations y = 3x - 9 and y = -2x + 16 are (x, y) = (5, 6)

The solution to given system of equations y = 5x + 22 and -6x - 4y = -10 are (x, y) = (-3, 7)

<em><u>Solution:</u></em>

Given that, we have to solve each system by substitution method

<em><u>Given system of equations are:</u></em>

y = 3x - 9 ----------- eqn 1

y = -2x + 16 -------- eqn 2

Substitution method is done by substituting eqn 2 in eqn 1

<em><u>Substitute the value of "y" from eqn 2 into eqn 1</u></em>

-2x + 16 = 3x - 9

Move the variables to one side and constants to other side

-2x - 3x = -9 - 16

Combine the like terms

-5x = -25

Cancel the negative sign on both sides of equation

5x = 25

x = \frac{25}{5} = 5

<h3>x = 5</h3>

<em><u>Substitute x = 5 in eqn 1</u></em>

y = 3(5) - 9

y = 15 - 9

<h3>y = 6</h3>

Thus solution to given system of equations are (x, y) = (5, 6)

<em><u>Given another system of equations are:</u></em>

y = 5x + 22 ----- eqn 1

-6x - 4y = -10 ------ eqn 2

Substitute eqn 1 in eqn 2

-6x - 4(5x + 22) = -10

-6x - 20x - 88 = -10

Move the variables to one side and constants to other side

-26x = -10 + 88

-26x = 78

<h3>x = -3</h3>

Substitute x = -3 in eqn 1

y = 5( - 3) + 22

y = -15 + 22

<h3>y = 7</h3>

Thus solution to given system of equations are (x, y) = (-3, 7)

4 0
3 years ago
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