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Jobisdone [24]
3 years ago
6

How do I simplify this?

Mathematics
1 answer:
ololo11 [35]3 years ago
5 0
Don't get too involved in the 'simplify' part just yet.
It doesn't mean make the problem easier.
It means that when you have the answer, make sure it's in simplest form.
That's why it says "Simplify your answer".
Notice that that's the 2nd instruction.
There's nothing to simplify until you have an answer.

The first thing you have to do here is find the area of a triangle.
Do you remember that formula ? 
I'll bet you do.
The area of a triangle is            (1/2) · (length of the base) · (height)  .

The height and the length of the base are given, right there
in the picture.  Take those, and use them to find the area.

Once you've multiplied the binomials, don't forget about the (1/2).

Finally, you have a trinomial for an answer.
Put the big sweet cherry on top:  Factor it !
I think that's the 'simplify' part.

You may be surprised.
I was when I worked it out just now ... just for you.
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What is the value of y?<br> A. 83<br> B. 89<br> C. 96<br> D. 97
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Answer:

<u>A. 83</u> is the correct option.

Step-by-step explanation:

all angles of a triangle are always equal to 180.

46 + 51 + y = 180

97 + y = 180

180 - 97 = <u>83</u>

Have a nice day! :-)

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Help me plssssssssssssssssssssssssssssss
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Step-by-step explanation:

1.) 1 gallon for $1.99

1/2 gallon for $0.98

1 gallon would be 0.98 × 2 = <em>1.96</em>

2 gallons for $3.98

1 gallon would be 3.98 ÷ 2 = 1.99

2.) 621 ÷ 4 = 155.25

He pays $155.25 each payment!

3.) V = s³

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V = 3³

3 × 3 × 3 = 27

V = 27 m³

4.) This is a rectangle; the area of a rectangle is l × b.

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5.) False

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Determine if the table is proportional or nonproportional
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The answer is nonproportional.

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If anyone knows about definite integrals for calculus then please I request help! I
kicyunya [14]

Answer:

\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)

General Formulas and Concepts:

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:                                                           \displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Integration

  • Integrals

Integration Rule [Fundamental Theorem of Calculus 1]:                                     \displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)

Integration Property [Multiplied Constant]:                                                         \displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx

U-Substitution

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx

<u>Step 2: Integrate Pt. 1</u>

<em>Identify variables for u-substitution.</em>

  1. Set <em>u</em>:                                                                                                             \displaystyle u = 4x^{-2}
  2. [<em>u</em>] Differentiate [Basic Power Rule, Derivative Properties]:                       \displaystyle du = \frac{-8}{x^3} \ dx
  3. [Bounds] Switch:                                                                                           \displaystyle \left \{ {{x = 9 ,\ u = 4(9)^{-2} = \frac{4}{81}} \atop {x = 5 ,\ u = 4(5)^{-2} = \frac{4}{25}}} \right.

<u>Step 3: Integrate Pt. 2</u>

  1. [Integral] Rewrite [Integration Property - Multiplied Constant]:                 \displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^9_5 {\frac{-8}{x^3}e^\big{4x^{-2}}} \, dx
  2. [Integral] U-Substitution:                                                                              \displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{4}{81}}_{\frac{4}{25}} {e^\big{u}} \, du
  3. [Integral] Exponential Integration:                                                               \displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}(e^\big{u}) \bigg| \limits^{\frac{4}{81}}_{\frac{4}{25}}
  4. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:           \displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8} \bigg( e^\Big{\frac{4}{81}} - e^\Big{\frac{4}{25}} \bigg)
  5. Simplify:                                                                                                         \displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)

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

Unit: Integration

4 0
2 years ago
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