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Mice21 [21]
3 years ago
11

3m=5(m+3)-3 How do I solve it

Mathematics
2 answers:
Mice21 [21]3 years ago
5 0

Simplifying

3m = 5(m + 3) + -3

Reorder the terms:

3m = 5(3 + m) + -3

3m = (3 * 5 + m * 5) + -3

3m = (15 + 5m) + -3

Reorder the terms:

3m = 15 + -3 + 5m

Combine like terms: 15 + -3 = 12

3m = 12 + 5m

Solving

3m = 12 + 5m

Solving for variable 'm'.

Move all terms containing m to the left, all other terms to the right.

Add '-5m' to each side of the equation.

3m + -5m = 12 + 5m + -5m

Combine like terms: 3m + -5m = -2m

-2m = 12 + 5m + -5m

Combine like terms: 5m + -5m = 0

-2m = 12 + 0

-2m = 12

Divide each side by '-2'.

m = -6

Simplifying

m = -6

FrozenT [24]3 years ago
4 0

Answer:

m = -6

Step-by-step explanation:

First you distribute 5 into (m+3) so it would 5m + 15

The equation now would be:

3m = 5m + 15 - 3

3m = 5m + 12

3m - 5m = 5m -5m + 12

-2m = 12

-2m/-2 = 12/-2

m = -6

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

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

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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

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