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3 years ago

Giving 21 points for answer!! Please help!!

Mathematics

2 answers:

Radda [10]3 years ago

6 0

Answer:

Helloooo. The bottom part is the only part of ur answer that's wrong. It is 8.

Step-by-step explanation:

I believe the other person that answered this shows this part correctly :)

nordsb [41]3 years ago

5 0

Answer:

The bottom question is wrong. You have to multiply 8 to get the amount earned. But the rest is correct.

Step-by-step explanation:

6.25 x 8 = 50

12.5 x 8 = 100

18.75 x 8 = 150

25 x 8 = 200

31.25 x 8 = 250

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Wats da anwser 4×70=4×7×_

Veseljchak [2.6K]

10 will be your answer

6 0

3 years ago

Read 2 more answers

Using the geometric mean and Pythagorean theorem, calculate the values of the missing sides. Round your answers to the thousandt

Pachacha [2.7K]

Answer:

a = 9.849

b = 20.25

c = 491.03

Step-by-step explanation:

By using Pythagoras theorem in the right triangle BDC,

(Hypotenuse)² = (Leg 1)² + (Leg 2)²

BC² = BD² + DC²

a² = 9² + 4²

a = $\sqrt{(81+16)}$

a = $\sqrt{97}$

a = 9.8489

a ≈ 9.849 units

By mean proportional theorem,

$\frac{\text{DC}}{\text{BD}}=\frac{\text{BD}}{\text{AD}}$

AD × DC = BD²

b × 4 = 9²

b = $\frac{81}{4}$

b = 20.25 units

BY Pythagoras theorem in ΔADB,

AB² = AD² + BD²

c² = b² + 9²

c² = (20.25)² + 9²

c² = 410.0625 + 81

c = 491.0625

c = 491. 063 units

6 0

3 years ago

Read each of the following situations and select ALL of the ones considered plagiarism.

mrs_skeptik [129]

Answer:

Ashley uses internal citation to give credit for someone else's idea

7 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:

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

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:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}$