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dangina [55]
3 years ago
12

PLEASE HELP FAST! Solve for Y in the following system of equations!

Mathematics
1 answer:
victus00 [196]3 years ago
7 0

Answer:

B.) 3

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

<u>Algebra I</u>

  • Solving systems of equations using substitution/elimination

Step-by-step explanation:

<u>Step 1: Define Systems</u>

x - y = -1

3x + 5y = 21

<u>Step 2: Rewrite Systems</u>

x - y = -1

  1. Add <em>y</em> to both sides:                    x = y - 1

<u>Step 3: Redefine Systems</u>

x = y - 1

3x + 5y = 21

<u>Step 4: Solve for </u><em><u>y</u></em>

<em>Substitution</em>

  1. Substitute in <em>x</em>:                              3(y - 1) + 5y = 21
  2. Distribute 3:                                   3y - 3 + 5y = 21
  3. Combine like terms:                     8y - 3 = 21
  4. Add 3 to both sides:                     8y = 24
  5. Divide 8 on both sides:                y = 3
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An engineer is planning a new water pipe installation. The circular pipe has a diameter of d=20\text{ cm}d=20 cmd, equals, 20, s
aivan3 [116]

Answer: The answer is 314.28 cm² (approx.).


Step-by-step explanation:  Given that an engineer is going to install a new water pipe. The diameter of this circular pipe is, d = 20 cm.

We need to find the area 'A' of the circular cross-section of the pipe.

Given, diameter of the circular section is

\textup{d}=20~\textup{cm}.

So, the radius of the circular cross-section will be

\textup{r}=\dfrac{\textup{d}}{2}=\dfrac{20}{2}=10~\textup{cm}.

Therefore, cross-sectional area of the pipe is

\textup{A}=\pi \textup{r}^2=\dfrac{22}{7}(10)^2=\dfrac{2200}{7}=314\dfrac{2}{7}=314.28~.~.~.~\textup{cm}^2.

Thus, the answer is 314.28 cm² (approx.).

4 0
4 years ago
What is the solution of VX-4+5=2?​
Katarina [22]

Answer:

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Step-by-step explanation:

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Answer:

x=3.5

Step-by-step explanation:

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8 0
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