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kompoz [17]
3 years ago
11

Help ASAP . i will give you points please !​

Mathematics
1 answer:
Pachacha [2.7K]3 years ago
5 0
It should be 150, 110+100=210 mines 360 equals 150
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Which point is a solution to the inequality shown in this graph? (3,-1) (-3,-3)
Ne4ueva [31]

Answer: try c (1,5)

Step-by-step explanation: hey eri-chan

5 0
3 years ago
If the office furniture costs $9,900 and it was three times as expensive as the office computer, what was the cost of the comput
Readme [11.4K]

Answer: 3,300

Step-by-step explanation:

9,900÷3

3 0
3 years ago
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Adjacent i need help understanding this
Cerrena [4.2K]

Answer:

The definition of adjacent in math can mean a couple of different things. For angles, adjacent angles would be any two angles that share a side. In other words, two angles right next to each other.

A more broad meaning is just any two geometric shapes that are next to each other in a figure.

8 0
3 years ago
Find the indicated limit, if it exists.
kondor19780726 [428]

Answer:

d) The limit does not exist

General Formulas and Concepts:

<u>Calculus</u>

Limits

  • Right-Side Limit:                                                                                             \displaystyle  \lim_{x \to c^+} f(x)
  • Left-Side Limit:                                                                                               \displaystyle  \lim_{x \to c^-} f(x)

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

Limit Property [Addition/Subtraction]:                                                                   \displaystyle \lim_{x \to c} [f(x) \pm g(x)] =  \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)

Step-by-step explanation:

*Note:

In order for a limit to exist, the right-side and left-side limits must equal each other.

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle f(x) = \left\{\begin{array}{ccc}5 - x,\ x < 5\\8,\ x = 5\\x + 3,\ x > 5\end{array}

<u>Step 2: Find Right-Side Limit</u>

  1. Substitute in function [Limit]:                                                                         \displaystyle  \lim_{x \to 5^+} 5 - x
  2. Evaluate limit [Limit Rule - Variable Direct Substitution]:                           \displaystyle  \lim_{x \to 5^+} 5 - x = 5 - 5 = 0

<u>Step 3: Find Left-Side Limit</u>

  1. Substitute in function [Limit]:                                                                         \displaystyle  \lim_{x \to 5^-} x + 3
  2. Evaluate limit [Limit Rule - Variable Direct Substitution]:                           \displaystyle  \lim_{x \to 5^+} x + 3 = 5 + 3 = 8

∴ Since  \displaystyle \lim_{x \to 5^+} f(x) \neq \lim_{x \to 5^-} f(x)  , then  \displaystyle \lim_{x \to 5} f(x) = DNE

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

Unit:  Limits

5 0
2 years ago
Use Simpson's Rule with n = 10 to approximate the area of the surface obtained by rotating the curve about the x-axis. Compare y
DiKsa [7]

The area of the surface is given exactly by the integral,

\displaystyle\pi\int_0^5\sqrt{1+(y'(x))^2}\,\mathrm dx

We have

y(x)=\dfrac15x^5\implies y'(x)=x^4

so the area is

\displaystyle\pi\int_0^5\sqrt{1+x^8}\,\mathrm dx

We split up the domain of integration into 10 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], ..., [4, 9/2], [9/2, 5]

where the left and right endpoints for the i-th subinterval are, respectively,

\ell_i=\dfrac{5-0}{10}(i-1)=\dfrac{i-1}2

r_i=\dfrac{5-0}{10}i=\dfrac i2

with midpoint

m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4

with 1\le i\le10.

Over each subinterval, we interpolate f(x)=\sqrt{1+x^8} with the quadratic polynomial,

p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}

Then

\displaystyle\int_0^5f(x)\,\mathrm dx\approx\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx

It turns out that the latter integral reduces significantly to

\displaystyle\int_0^5f(x)\,\mathrm dx\approx\frac56\left(f(0)+4f\left(\frac{0+5}2\right)+f(5)\right)=\frac56\left(1+\sqrt{390,626}+\dfrac{\sqrt{390,881}}4\right)

which is about 651.918, so that the area is approximately 651.918\pi\approx\boxed{2048}.

Compare this to actual value of the integral, which is closer to 1967.

4 0
3 years ago
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