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Lilit [14]
2 years ago
7

-712 ÷ 17= help please

Mathematics
1 answer:
dusya [7]2 years ago
7 0

Answer:

-41.88

Step-by-step explanation:

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We are asked to factor out the expression:

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2 x ( 2 x^2 - 5)

Therefore, please select the first answer option they give you in the list.

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1 year ago
Cherie measures and records the lengths and sizes of the same style of a sandal found at a shoe store. A 2-column table with 5 r
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Step-by-step explanation:

6 0
3 years ago
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What is the quotient of StartFraction 7 Superscript negative 1 Baseline Over 7 Superscript negative 2 Baseline EndFraction? Star
Komok [63]

The quotient of the number given number  7 Superscript negative 1 Baseline Over 7 Superscript negative 2 Baseline is 7.

<h3>What is the quotient?</h3>

Quotient is the resultant number which is obtain by dividing a number with another. Let a number a is divided by number b. Then the quotient of these two number will be,

q=\dfrac{a}{b}

Here, (<em>a, b</em>) are the real numbers.

The number StartFraction 7 Superscript negative 1 Baseline Over 7 Superscript negative 2 Baseline EndFraction, given can be written as,

\dfrac{7^{-1}}{7^{-2}}

Let the quotient of this division is n. Therefore,

n=\dfrac{7^{-1}}{7^{-2}}

A number in numerator of a fraction with negative exponent can be written in the denominator with the same but positive exponent and vise versa. Therefore,

n=\dfrac{7^{2}}{7^{1}}\\n=7

Hence, the quotient of the number given number  7 Superscript negative 1 Baseline Over 7 Superscript negative 2 Baseline is 7.

Learn more about the quotient here;

brainly.com/question/673545

7 0
2 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:

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

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

Basic Power Rule:

  1. f(x) = cxⁿ
  2. 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:

<u>Step 1: Define</u>

<em>Identify</em>

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

<u>Step 2: Integrate Pt. 1</u>

<em>Identify variables for u-substitution.</em>

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

<u>Step 3: Integrate Pt. 2</u>

  1. [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
  2. [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
  3. [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}}
  4. 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)
  5. 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

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