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3 years ago

Which answer describes the calculations that could be used to solve this problem?

Mathematics

2 answers:

Ber [7]3 years ago

7 0

What you do is add 8+8+8+8+8+8+8 so the simple form is 8x7, and then you subtract 100 and you should get your answer.

faltersainse [42]3 years ago

4 0

D. Since we know that 1 ticket = $8 we would multiply to figure out how much 7 tickets were. (7x$8). And to figure out his change you would subtract the ticket amount from $100 :) Hope this helps

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Factor the trinomial x^2 - 2x - 8

ValentinkaMS [17]

Well what two numbers multiply to get -8 and add to get -2?
Well, to get -8, we either have (-1,8),(-2,4),(-4,2),(-8,1)
8-1=7
-2+4=2
2-4=-2
1-8=-7
Thus it should be (-4,2),
This means we should get (X-4)(X+2)

7 0

2 years ago

Sam has 3x + 4 dollars saved. Tessa has saved 7x + 16 dollars. Together they

Shalnov [3]

Step-by-step explanation:

3x + 4 + 7x + 16 = 110

10x = 90

x = 9

therefore, Sam = $31 and Tessa = $79

Topic: Algebraic Equations

If you like to venture further, do check out my insta (learntionary) where I constantly post math tips! Thanks!

8 0

3 years ago

Read 2 more answers

-45×68 what is the answer

Marianna [84]

Answer: -3060

68
-45
--------
340
272x <-- the x is in place of a zero
--------
3060

times the negative: -3060

4 0

2 years ago

When renting a limo for prom, the number of people varies inversely with the cost per person. Originally there were 4 people and

ioda

Answer:The new cost per person= $19 when there are 12 people

Step-by-step explanation:

Step 1

Let Number of people be rep as N

And Cost per person = C

such that the relationship of number of people which varies inversely with the cost per person is given as

N∝ 1/C

Introducing constant of proportionality, k, we have that

N= K/C

such that when there were 4 people the cost per person was $57, K would be

N= K/C

4 = K /57

k= 57 X 4

k=228

Step 2

Therefore If the number of people changed to 12, cost per person would be

N= K/C

12= 228 / C

C= 228/12

C= 19

The new cost per person= $19 when there are 12 people

4 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:

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \boxed{ \frac{1}{4} }$

General Formulas and Concepts:

Calculus

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]:

f(x) = cxⁿ
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:

Step 1: Define

Identify given limit.

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

Step 2: Find Limit

Let's start out by directly evaluating the limit:

[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \frac{\sqrt{3 + 1} - 2}{3 - 3}$
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:

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

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

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

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

Unit: Limits

3 0

1 year ago