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3 years ago

Sinx•Tanx•Cotx•Cscx =

Mathematics

1 answer:

nekit [7.7K]3 years ago

8 0

Answer:

the answer of this question is " 1 "

Step-by-step explanation:

tanx = sinx/cosx

cotx = cosx/sinx

cscx = 1/sinx

after putting them all in the given equation

sinx.(sinx/cosx).(cosx/sinx).(1/sinx)

they all are canceled as all are being multiplied with their reciprocals so answer is " 1 "

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The mean height of women in a country (ages 20-29) is 63.9 inches. A random sample of 65 women in this age group is selected. Wh

anzhelika [568]

Answer: The mean height of women in a country ages 20 29

The mean height of women in a country (ages 20-29) is 63.7 inches.

4 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

2 years ago

Here are my points idc anymore

RSB [31]

Answer:

Thanks for the free points

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Construct a quadratic polynomial whose zeroes are negatives of the zeroes of the

sp2606 [1]

Given:

The given quadratic polynomial is :

$x^2-x-12$

To find:

The quadratic polynomial whose zeroes are negatives of the zeroes of the given polynomial.

Solution:

We have,

$x^2-x-12$

Equate the polynomial with 0 to find the zeroes.

$x^2-x-12=0$

Splitting the middle term, we get

$x^2-4x+3x-12=0$

$x(x-4)+3(x-4)=0$

$(x+3)(x-4)=0$

$x=-3,4$

The zeroes of the given polynomial are -3 and 4.

The zeroes of a quadratic polynomial are negatives of the zeroes of the given polynomial. So, the zeroes of the required polynomial are 3 and -4.

A quadratic polynomial is defined as:

$x^2-(\text{Sum of zeroes})x+\text{Product of zeroes}$

$x^2-(3+(-4))x+(3)(-4)$

$x^2-(-1)x+(-12)$

$x^2+x-12$

Therefore, the required polynomial is $x^2+x-12$ .

4 0

3 years ago

Please answer ASAP!!!

drek231 [11]

Hi there!
------------------------
question - A 36-foot flagpole casts a 10 3/4 foot shadow, while the hospital nearby casts a 37 5/8 foot shadow. Find the height of the hospital
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if you correctly work this out the answer is:
The height of the hospital is 35 3/4 feet
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Have a nice day!

6 0

3 years ago