Round each decimal to the nearest one then subtract $3.85-$2.17

Which is greater 3.75 3.750

guapka [62]

Answer:

None

Step-by-step explanation:

They are the same number

5 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

What are the coordinates of the endpoints of the midsegment for DEF that is parallel to DE

Nutka1998 [239]

Answer:

$\left(\dfrac{x_D+x_F}{2},\dfrac{y_D+y_F}{2}\right),$ $\left(\dfrac{x_E+x_F}{2},\dfrac{y_E+y_F}{2}\right).$

Step-by-step explanation:

Let points D, E and F have coordinates $(x_D,y_D),\ (x_E,y_E)$ and $(x_F,y_F).$

1. Midpoint M of segment DF has coordinates

$\left(\dfrac{x_D+x_F}{2},\dfrac{y_D+y_F}{2}\right).$

2. Midpoint N of segment EF has coordinates

$\left(\dfrac{x_E+x_F}{2},\dfrac{y_E+y_F}{2}\right).$

3. By the triangle midline theorem, midline MN is parallel to the side DE of the triangle DEF, then points M and N are endpoints of the midsegment for DEF that is parallel to DE.

6 0

3 years ago

Please answer the 2 question l will mark as brainiest

N76 [4]

Answer:

1. A=πr2=π·4.62≈66.4761

A≈66.48

2. A=πr2=π·22≈12.56637

A≈12.57

please mark as brainiest

3 0

3 years ago

Read 2 more answers

Yasmin purchased 6 heads of cabbage that each weighed pounds. How much did the cabbage weigh all together?

MakcuM [25]

The weight of each cabbage is not given.
I'll assume that each cabbage weighs p pounds.

Now, we know that Yasmin bought 6 cabbage each weighing p pounds. To know the total weight, we will simply multiply the number of cabbages by the weight of each as follows:
Total weight = 6 * p = 6p pounds

Hope this helps :)

5 0

3 years ago

Read 2 more answers