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2 years ago

Find the 5th term of the GP is 48 and it 8th term is 384. find the first term and the common

Mathematics

1 answer:

rewona [7]2 years ago

7 0

Answer:

$\text{1st term} = 3\\\\\text{Common ratio} = 2$

Step by step explanation:

$\text{Given that,}\\\\\text{5th term} = ar^{5-1} = ar^4 = 48~~~~~~~~...(i)\\ \\\text{8th term} = ar^{8-1} = ar^7 = 384~~~~~~~...(ii)\\\\\text{1st term,}~ a = ?\\\\\text{Common ratio,}~ r = ?\\\\(ii) \div (i):\\\\~~~~~~\dfrac{ar^7}{ar^4}=\dfrac{384}{48}\\\\\implies r^{7-4} = 8\\\\\implies r^3 = 8\\\\\implies r^3 = 2^3\\\\\implies r =2\\\\\text{Substitute r = 2 in eq (i):}\\\\~~~~~~a\cdot 2^4 = 48\\\\\implies 16a =48\\\\\implies a = \dfrac{48}{16}\\\\\implies a = 3\\$

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Help plz answer quick

Inessa [10]

Answer:

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Step-by-step explanation:

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8 0

3 years ago

If anyone knows about definite integrals for calculus then please I request help! I

kicyunya [14]

Answer:

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals

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

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

U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

Set u: $\displaystyle u = 4x^{-2}$
[u] Differentiate [Basic Power Rule, Derivative Properties]: $\displaystyle du = \frac{-8}{x^3} \ dx$
[Bounds] Switch: $\displaystyle \left \{ {{x = 9 ,\ u = 4(9)^{-2} = \frac{4}{81}} \atop {x = 5 ,\ u = 4(5)^{-2} = \frac{4}{25}}} \right.$

Step 3: Integrate Pt. 2

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^9_5 {\frac{-8}{x^3}e^\big{4x^{-2}}} \, dx$
[Integral] U-Substitution: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{4}{81}}_{\frac{4}{25}} {e^\big{u}} \, du$
[Integral] Exponential Integration: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}(e^\big{u}) \bigg| \limits^{\frac{4}{81}}_{\frac{4}{25}}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8} \bigg( e^\Big{\frac{4}{81}} - e^\Big{\frac{4}{25}} \bigg)$
Simplify: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

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

Unit: Integration

4 0

3 years ago

FIND THE SUM AND THE DIFFERENCE FOR EACH PAIR OF POLYNOMIALS please show work

Pepsi [2]

Answer:

12a. Addition = 7y⁵ + 7y³ – 6y – 5

12b. Subtraction = – y⁵ – 3y³ + 8y + 11

13a. Addition = 5x⁴ + x³ – 6x² + 8x + 17

13b. Subtraction = – x⁴ – 9x³ –6x² + 8x + 3

14a. Addition = – 2b⁴ + b³ – 11b² – 1

14b. Subtraction = 4b⁴ + 7b³ – b² + 8b – 1

15a. Addition = m⁵ + m⁴ + m³ + m²

15b. Subtraction = m⁵ – m⁴ – m³ – m² – 10

16a. Addition = 4x⁴ + 12x³ + 18x² + 16x + 4

16b. Subtraction = 4x⁴ – 4x³ + 14x² – 16x + 4

Step-by-step explanation:

12a. Addition

.. 3y⁵ + 2y³ + y + 3

+ (4y⁵ + 5y³– 7y – 8)

————————————

= 7y⁵ + 7y³ – 6y – 5

12b. Subtraction

.. 3y⁵ + 2y³ + y + 3

– (4y⁵ + 5y³– 7y – 8)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= – y⁵ – 3y³ + 8y + 11

13a. Addition

.. 2x⁴ – 4x³ – 6x² + 8x + 10

+ (3x⁴ + 5x³ +...................+ 7)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= 5x⁴ + x³ – 6x² + 8x + 17

13b. Subtraction

.. 2x⁴ – 4x³ – 6x² + 8x + 10

– (3x⁴ + 5x³ +...................+ 7)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= – x⁴ – 9x³ – 6x² + 8x + 3

14a. Addition

...... b⁴ + 4b³ – 6b² + 4b – 1

+ (– 3b⁴ – 3b³ – 5b² – 4b)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= – 2b⁴ + b³ – 11b² – 1

14b. Subtraction

...... b⁴ + 4b³ – 6b² + 4b – 1

– (– 3b⁴ – 3b³ – 5b² – 4b)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= 4b⁴ + 7b³ – b² + 8b – 1

15a. Addition

... m⁵ + m³ – 5

+ (m⁴ + m² + 5)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= m⁵ + m⁴ + m³ + m²

15b. Subtraction

... m⁵ + m³ – 5

– (m⁴ + m² + 5)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= m⁵ – m⁴ – m³ – m² – 10

16a. Addition

.... 4x⁴ + 8x³ + 16x² + 4

+ (4x³ + 2x² + 16x)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= 4x⁴ + 12x³ + 18x² + 16x + 4

16b. Subtraction

.... 4x⁴ + 8x³ + 16x² + 4

– (4x³ + 2x² + 16x)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

= 4x⁴ – 4x³ + 14x² – 16x + 4

3 0

3 years ago

When the hour hand moves from 3 pm to 6 pm, through how many degrees does it move?

taurus [48]

I think it might be 90 degrees

6 0

3 years ago

(05.07 MC)

True [87]

I think 10.24 sq. ft

4 0

3 years ago

Read 2 more answers

Find the 5th term of the GP is 48 and it 8th term is 384. find the first term and the common​

Find the 5th term of the GP is 48 and it 8th term is 384. find the first term and the common