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Zanzabum
3 years ago
7

How can i solve the following 2(x + 3) = x - 4

Mathematics
2 answers:
julsineya [31]3 years ago
4 0

Answer:

x=-10

Step-by-step explanation:

2(x+3)=x-4

2*x+2*3=x-4

2x+6=x-4

2x-x=-4-6

x=-10

Stels [109]3 years ago
4 0

Answer:

x =  - 10

Step-by-step explanation:

Let's solve:

2(x+3)=x−4

Step 1: Simplify both sides of the equation.

2(x+3)=x−4 \\ (2)(x)+(2)(3)=x+−4(Distribute) \\ 2x+6=x+−4 \\ 2x+6=x−4

Step 2: Subtract x from both sides.

2x+6−x=x−4−x \\ x+6=−4

Step 3: Subtract 6 from both sides.

x+6−6=−4−6 \\ x=−10

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Determine the area enclosed by y=2x+3, the x-axis and the ordinates x=3 and x=4​
jok3333 [9.3K]

Answer:

\displaystyle \int\limits^4_3 {2x + 3} \, dx = 10

General Formulas and Concepts:
<u>Calculus</u>

Integration

  • Integrals

Integration Rule [Reverse Power Rule]:                                                           \displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C

Integration Rule [Fundamental Theorem of Calculus 1]:                                 \displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)

Integration Property [Multiplied Constant]:                                                     \displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx

Integration Property [Addition/Subtraction]:                                                   \displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx

Area of a Region Formula:                                                                               \displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify.</em>

y = 2x + 3

<em>x</em>-interval [3, 4]

<em>x</em>-axis

<em>See attachment for graph.</em>

<u>Step 2: Find Area</u>

  1. Substitute in variables [Area of a Region Formula]:                               \displaystyle A = \int\limits^4_3 {2x + 3} \, dx
  2. [Integral] Rewrite [Integration Property - Addition/Subtraction]:           \displaystyle A = \int\limits^4_3 {2x} \, dx + \int\limits^4_3 {3} \, dx
  3. [Integrals] Rewrite [Integration Property - Multiplied Constant]:           \displaystyle A = 2 \int\limits^4_3 {x} \, dx + 3 \int\limits^4_3 {} \, dx
  4. [Integrals] Integrate [Integration Rule - Reverse Power Rule]:               \displaystyle A = 2 \bigg( \frac{x^2}{2} \bigg) \bigg| \limits^4_3 + 3(x) \bigg| \limits^4_3
  5. [Integrals] Integrate [Integration Rule - FTC 1]:                                       \displaystyle A = 2 \bigg( \frac{7}{2} \bigg) + 3(1)
  6. Simplify:                                                                                                     \displaystyle A = 10

∴ the area bounded by the region y = 2x + 3, x-axis, and the coordinates x = 3 and x = 4 is equal to 10.

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Learn more about integration: brainly.com/question/26401241

Learn more about calculus: brainly.com/question/20197752

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

Unit: Integration

3 0
2 years ago
What is h less than 30
horrorfan [7]
H< 30

That is your inequality. 
8 0
3 years ago
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