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Rudik [331]
3 years ago
13

The number of Roberto’s baseball cards is ¾ the number of David’s cards If Roberto gives ½ of his cards to David, what will be t

he ratio of Roberto’s cards to David’s cards? please help me. I need it to step by step equations!!!!
Mathematics
2 answers:
inn [45]3 years ago
6 0

Answer:

1/4, 0.25

Step-by-step explanation:

PtichkaEL [24]3 years ago
4 0

Answer:

3/11

Step-by-step explanation:

Let Roberto's cards be x.

Let David's cards be y

x = (3/4) y

Roberto gives half his cards away which is 1/2*(3/4) = 3/8

David's new amount after the gift = 1/2(3/4)y + y = 3/8y+y=3y/8+8y/8 = 11y/8

So the ratio is Roberto / David = (3/8) // (11/8)

Roberto to David = 3/8 // 11.8 which you now invert and multiply the denominator

Roberto to David = 3/8 * 8/11 = 3/11

Done on my cell.

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KatRina [158]

Answer:

The answer is 32.63

Step-by-step explanation:

4 0
3 years ago
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Which is equal to 8^-10/8^5 ?
makkiz [27]
8^{-10} can be converted into \frac{1}{ 8^{10} } , and applying this to 8^{-5} can see that

\frac{ \frac{1}{ 8^{10} } }{ \frac{1}{ 8^{5} } }

This gives us 

\frac{ 8^{5} }{ 8^{10} }  =  \frac{1}{ 8^{5} } = 8^{-5}
8 0
2 years ago
In a bag full of small marbles, 1/4 of the marbles are green 1/8 of the marbles are blue and 1/12 of the marbles are
Alik [6]

Step-by-step explanation:

white=26

1/4+1/8+1/12=6/24+3/24+2/24=11/24=(green,blue and yellow marbles)

24-11=13/24=white

26/13=2=1/24

blue=3/24

2.3=6

answer 6

8 0
3 years ago
What is Limit of StartFraction StartRoot x + 1 EndRoot minus 2 Over x minus 3 EndFraction as x approaches 3?
scoray [572]

Answer:

<u />\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \boxed{ \frac{1}{4} }

General Formulas and Concepts:

<u>Calculus</u>

Limits

Limit Rule [Variable Direct Substitution]:
\displaystyle \lim_{x \to c} x = c

Special Limit Rule [L’Hopital’s Rule]:
\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Addition/Subtraction]:
\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]
Derivative Rule [Basic Power Rule]:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:
\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify given limit</em>.

\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3}

<u>Step 2: Find Limit</u>

Let's start out by <em>directly</em> evaluating the limit:

  1. [Limit] Apply Limit Rule [Variable Direct Substitution]:
    \displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \frac{\sqrt{3 + 1} - 2}{3 - 3}
  2. Evaluate:
    \displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \frac{\sqrt{3 + 1} - 2}{3 - 3} \\& = \frac{0}{0} \leftarrow \\\end{aligned}

When we do evaluate the limit directly, we end up with an indeterminant form. We can now use L' Hopital's Rule to simply the limit:

  1. [Limit] Apply Limit Rule [L' Hopital's Rule]:
    \displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\\end{aligned}
  2. [Limit] Differentiate [Derivative Rules and Properties]:
    \displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \leftarrow \\\end{aligned}
  3. [Limit] Apply Limit Rule [Variable Direct Substitution]:
    \displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \leftarrow \\\end{aligned}
  4. Evaluate:
    \displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}

∴ we have <em>evaluated</em> the given limit.

___

Learn more about limits: brainly.com/question/27807253

Learn more about Calculus: brainly.com/question/27805589

___

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

Unit: Limits

3 0
1 year ago
Which figure is described below?
IRINA_888 [86]

Answer:

Option A.

Step-by-step explanation:

Circle contains all points in a plane that are equidistant from a point, i.e., center of the circle.

The locus of points 9 units from the  point (-1,3) on the coordinate plane.

It means, the figure represents the set of all points which are 9 units from the  point (-1,3).

So, the given describes a circle with of 9 units and center at (-1,3).

Therefore, the correct option is A.

5 0
3 years ago
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