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3 years ago

Divide and round to the nearest tenth 3.5 divided by 2.29

Mathematics

1 answer:

Jlenok [28]3 years ago

3 0

1.53 is your answer
Hope this helps!

You might be interested in

Using this formula, how many sides does a polygon have if the sum of the interior angles is 1,260°? Round to the nearest whole n

nikdorinn [45]

Answer:

9 Sides

Step-by-step explanation:

According to the Question,

Given That, The sum of the interior angles(s), in an n-sided polygon can be determined using the formula s=180(n−2), where n is the number of sides

Therefore, The sides do a polygon have if the sum of the interior angles is 1,260°.

Put The Values in Formula, we get

1260=180(n−2)

n-2 = 7

n=9 (9 Sides)

6 0

3 years ago

Assume the rate of inflation is 5% per year for the next 2 years. What will be the cost of goods 2 years from now, adjusted fo

kirza4 [7]

The equation for inflation is
A = P*(1+r)^t
which is an exponential growth equation (if r > 0). If r < 0, then we have deflation.

where...
A = final price after t years
P = initial starting price
r = rate of inflation in decimal form
t = number of years

In this case,
A = unknown (we're solving for this)
P = 280 is the starting price
r = 0.05 is the decimal form of 5%
t = 2 years

We will plug these three pieces of info into the formula to get...
A = P*(1+r)^t
A = 280*(1+0.05)^2
A = 280*(1.05)^2
A = 280*(1.1025)
A = 308.70

Answer: 308.70 dollars

7 0

4 years ago

What are the solutions to the quadratic equation -2x^2 + 6x + 3 = 0?

mario62 [17]

For this, we will be using the quadratic formula, which is $x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}$ , with a=x^2 coefficient, b=x coefficient, and c = constant. Our equation will look like this: $x=\frac{-6+/-\sqrt{6^2-4*(-2)*3}}{2*(-2)}$

Firstly, solve the multiplications and the exponents: $x=\frac{-6+/-\sqrt{36+24}}{-4}$

Next, do the addition: $x=\frac{-6+/-\sqrt{60}}{-4}$

Next, your equation will be split into two: $x=\frac{-6+\sqrt{60}}{-4},\frac{-6-\sqrt{60}}{-4}$ . Solve them separately, and your answer will be $x=-0.436,3.436$

5 0

4 years ago

a wire of length 200 cm is cut into two parts and each part is bent to form a square.If the area of the larger square is 9 times

diamong [38]

Answer:

The perimeter of the larger square is $150cm.$

Step-by-step explanation:

First of all, let one part of it be " $x$ "

and the other part of it " $200-x$ "

Now to solve this :

The area of the square = $L^2$

The area of one part of the square = $x^2$

The area of the other part of the square = $(200-x)^2$

$9x^2= (200-x)^2\\9x^2=40000-400x+x^2$

Now, add, $-4x^2$ to both the sides :

$9x^2-9x^2=40000-400x+x^2-9x^2\\-1(8x^2+400x-40000)=0$

Now, take out the "8" which is common :

$-8(x^2+50x-5000)=0$

Now, divide it by $-8$ :

$x^2+100x-50x-5000=0\\x(x+100)-50(x+100)=0\\(x+100)(x-50)=0\\$

$x=-100$ or $50$

So, now we know that :

One of the part is = $50cm$

And the other part is = $150cm$

Thus the perimeter of the larger square is = $150cm$

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

∴ 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

2 years ago