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grigory [225]
3 years ago
9

Which ratio represents the value of sin Q

Mathematics
1 answer:
Volgvan3 years ago
4 0

We know that :

\bigstar \ \ \boxed{\sf{sin\theta = \dfrac{Opposite \ Side}{Hypotenuse}}}

From the figure, We can notice that :

↔   Opposite side of Angle Q is 9 units

↔   Hypotenuse is 15 units

\sf{\implies sinQ = \dfrac{9}{15}}

\sf{\implies sinQ = \dfrac{3}{5}}

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Answer:

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7 0
3 years ago
Factorise the following:<br> a) x + 10x + 16
sertanlavr [38]

Answer:

<h3><em>11x + 16</em></h3>

Step-by-step explanation:

x + 10x = 11x

= 11x + 16

4 0
3 years ago
Hannah buys 3 pack of 36 exposure film and 2 packs of 24 exposure film.She uses 8 rolls of film.How many rolls does she have lef
xeze [42]
3 times 36= 108
2 times 24= 48
108+48= 156
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3 0
3 years ago
3X + 2 = 17 how do I show my work for this?
SCORPION-xisa [38]

Answer:

x = 5

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

3x + 2 = 17

<u>Step 2: Solve for </u><em><u>x</u></em>

  1. [Subtraction Property of Equality] Subtract 2 on both sides:                        3x = 15
  2. [Division Property of Equality] Divide 3 on both sides:                                  x = 5
8 0
3 years ago
Read 2 more answers
The probability that a random student at Elmville College is a freshman is 0.3; a sophomore, 0.25; and a junior or senior, 0.45.
Andre45 [30]
Let L_1 denote the event that a student is a freshman, L_2 a sophomore, L_3 a junior, and L_4 a senior.

Let E denote the event that a student majors in engineering. By the law of total probability,

\mathbb P(E)=\mathbb P(E\cap L_1)+\mathbb P(E\cap L_2)+\mathbb P(E\cap L_3)+\mathbb P(E\cap L_4)

By the definition of conditional probability, we can expand each of these intersection probabilities to get

\mathbb P(E)=\mathbb P(E\mid L_1)\mathbb P(L_1)+\mathbb P(E\mid L_2)\mathbb P(L_2)+\mathbb P(E\mid L_3)\mathbb P(L_3)+\mathbb P(E\mid L_4)\mathbb P(L_4)

\mathbb P(E)=0.15\cdot0.3+0.2\cdot0.25+2(0.3\cdot0.45)=0.365
8 0
3 years ago
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