All categories

2 years ago

carlos cut 5/12 of a yard of cloth into 5 pieces of equal length . what was the length. of each piece of cloth ?

Mathematics

2 answers:

PolarNik [594]2 years ago

6 0

Answer:

the answer is a

Step-by-step explanation:

Travka [436]2 years ago

4 0

Answer:

The length of each piece of cloth =1/12 yard

Step-by-step explanation:

Number of pieces = 5

Carlos cut 5/12 of a yard of cloth

Length of each piece = ?

To find the length of each piece simply divide 5/12 by 5.

Length of each piece of cloth = 5/12/5

Length of each piece of cloth =5/12 * 1/5

Length of each piece of cloth=1/12

Therefore the length of each piece of cloth =1/12 yard....

You might be interested in

The inequality x ≥ -3 is satisfied by point (0, -2). TrueFalse

tino4ka555 [31]

Answer:

True

On a graph, in (0,-2) , the x position is to the right of -3

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

please help i rly need it my math teacher gave me over 30 assignments and a project that are all due tomorrow today and i want t

Sedaia [141]

Answer:

x= -14

Step-by-step explanation:

-14 subtracted by 2 is -16, since its absolute it means its positive, meaning its 16 greater than 14, I may be wrong so just to let you know. Depends on how the teacher wants it to be.

8 0

2 years ago

g What is the probability of being dealt exactly three of a kind (like three kings or three 7’s, etc.) in a five card hand from

Elena L [17]

Answer: 0.02257

Step-by-step explanation:

Given : Total cards in a deck = 52

Number of ways to select any 5 cards : $^{52}C_5$

Since , there are total 13 kinds of card (includes Numbers from 2 to 9 and Ace , king, queen and jack).

Of each kind , there are 4 cards.

Number of ways to select three cards in a five card hand of a single kind : $^{4}C_3\times^{48}C_2$

Number of ways to select three cards in a five card hand of a exactly three of a kind : $13\times^{4}C_3\times^{48}C_2$

Now , the required probability = $\dfrac{13\times^{4}C_3\times^{48}C_2}{^{52}C_5}$

$=\dfrac{13\times4\times\dfrac{48!}{2!46!}}{\dfrac{52!}{5!47!}}\\\\\\=\dfrac{58656}{2598960}$

$=0.022569027611\approx0.02257$

∴ The probability of being dealt exactly three of a kind (like three kings or three 7’s, etc.) in a five card hand from a deck of 52 cards= 0.02257

8 0

2 years ago

The model represents an equation. What value of x makes the equation true?

Citrus2011 [14]

Answer:

the normal x is three and the red x is negative three

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

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