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Vedmedyk [2.9K]
1 year ago
6

Help me asap! I will give you marks

Mathematics
1 answer:
Law Incorporation [45]1 year ago
6 0

Recall the binomial theorem.

(a+b)^n = \displaystyle \sum_{k=0}^n \binom nk a^{n-k} b^k

1. The binomial expansion of \left(1+\frac x3\right)^7 is

\left(1 + \dfrac x3\right)^7 = \displaystyle\sum_{k=0}^7 \binom 7k 1^{7-k} \left(\frac x3\right)^k = \sum_{k=0}^7 \binom 7k \frac{x^k}{3^k}

Observe that

k = 1 \implies \dbinom 71 \left(\dfrac x3\right)^1 = \dfrac73 x

k = 2 \implies \dbinom 72 \left(\dfrac x3\right)^2 = \dfrac73 x^2

When we multiply these by 8-9x,

• 8 and \frac73 x^2 combine to make \frac{56}3 x^2

• -9x and \frac73 x combine to make -\frac{63}3 x^2 = -21x^2

and the sum of these terms is

\dfrac{56}3 x^2 - 21x^2 = \boxed{-\dfrac73 x^2}

2. The binomial expansion is

\left(2a - \dfrac b2\right)^8 = \displaystyle \sum_{k=0}^8 \binom 8k (2a)^{8-k} \left(-\frac b2\right)^k = \sum_{k=0}^8 \binom 8k 2^{8-2k} a^{8-k} b^k

We get the a^6b^2 term when k=2 :

k=2 \implies \dbinom 82 2^{8-2\cdot2} a^{8-2} b^2 = 28 \cdot2^4 a^6 b^2 = \boxed{448} \, a^6b^2

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What's the area of a circle with diameter 18 units? Question 8 options: A) 9π units2 B) 81π units2 C) π units2 D) 18π units2
Anna [14]

Answer:

B) 81π units²

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
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Step-by-step explanation:

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Diameter <em>d</em> = 18 units

<u>Step 2: Manipulate Variables</u>

Radius <em>r</em> = 18 units/2 = 9 units

<u>Step 3: Find Area</u>

  1. Substitute in <em>r</em> [Area of a Circle Formula]:                                                       A = π(9 units)²
  2. [Area] Evaluate exponents:                                                                              A = π(81 units²)
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Solve completely (all missing sides and all missing angles) using trig ratios or laws. Round to the nearest tenth.
hodyreva [135]

Answer:

Step-by-step explanation:

3. Use Cosine law to find the  length of the unknown side (PR)

q² = p² +  r² - 2prCos Q

q is the opposite side of ∠Q;  

p is the opposite side of ∠P; p = 33

r is the opposite sides of ∠R ; r = 67

q² = 33² + 67² -  2* 33*67 Cos 19°

    = 1089 + 4489 - 4422 * 0.95

     = 1089 + 4489 - 4200.9

    = 1377.1

q = √1377.1

q = 37.1

PR = 37.1

To find the angle use law of sin

\sf \dfrac{q}{Sin \ Q} =\dfrac{p}{Sin \ P}\\\\\\\dfrac{37.1}{Sin \ 19}=\dfrac{33}{Sin \ P} \\\\\\\dfrac{37.1}{0.33}=\dfrac{33}{Sin \ P}\\\\\\37.1* Sin \ P = 33 *0.33\\\\Sin \ P =\dfrac{33*0.33}{37.1}

Sin P = 0.3

P = Sin^{-1} 0.3\\

P = 17.5°

∠R = 180 - (19 + 17.5)

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