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Nutka1998 [239]
3 years ago
7

Elise won 40 lollipops and the school fair. she gave 2 to every student in her math class and has at least 7 left. what is the m

aximum number of students in her class?
Mathematics
2 answers:
geniusboy [140]3 years ago
8 0
40-7= 33
33/2= 16.5
The maximum would be 17 students
Alchen [17]3 years ago
3 0
40 lollipops minus at least 7 = 33 lollipops33 lollipops divided by 2 = 16.5 classmatesElisa has 16 classmates who were given 2 lollipops, so she has 8 leftover of the 40 she won playing basketball.

Hope this helps  :)
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nika2105 [10]

Answer:

Step-by-step explanation:

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3 years ago
Solve -1 < x + 1 ≤ 6
andrey2020 [161]

Answer:

-2<x≤5

Step-by-step :

-1<x+1≤6 So fisrt what every we do to one side we must do to the other. In this case it's a bit different since we are dealing with inequalities.

-1<x+1≤6 I would start off by isolating x in the middle.

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-1     -1   -1

Now your equation should look like this:

-2<x≤5  Now there is really nothing much we can do here since we were just trying to get x by its self.

Answer : -2<x≤5

7 0
2 years ago
the head of a nail is circular , as shown the head of this nail has a diameter of 6 millimeters what measurment is closet to the
Nikitich [7]

Answer:

Area of nail's head  = 28.26 millimeter² (Approx.)

Step-by-step explanation:

Given:

Head of nail is circular

Diameter of nail's head = 6 millimeter

Find:

Area of nail's head

Computation:

Radius of nail's head = Diameter / 2

Radius of nail's head = 6 / 2

Radius of nail's head = 3 millimeter

Area of circle = πr²

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Area of nail's head  = (22/7)(3)²

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5 0
2 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
2 years ago
In a triangle, two angles have measures of 45 degrees and 85 degrees. What is the measure of the third angle in this triangle?
Blizzard [7]

Answer:

50 degrees

Step-by-step explanation:

All three angles in a triangle add up to 180 degrees. If you know the measure of two angles, add them together and subtract the sum from 180.  

45+85=130.  Then, subtract 130 from the total, or 180.   180-130=50. The measure of the third angle of the triangle is 50 degrees.

5 0
3 years ago
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