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gulaghasi [49]
3 years ago
7

The size of a TV is measured by its diagonal. You want to buy a TV that has a 32

Mathematics
1 answer:
ANEK [815]3 years ago
7 0

The width of the TV is 27.9 inches

Step-by-step explanation:

Given,

Diagonal of TV = 32 inch

Height of TV = 15.7 inch

The diagonal makes the TV into a right triangle,

Therefore,

Diagonal = hypotenuse = c = 32 inch

Height = perpendicular = a = 15.7 inch

Width = base = b

Using pythagoras theorem

a^2+b^2=c^2\\(15.7)^2+b^2=(32)^2\\246.49+b^2=1024\\b^2=1024-246.49\\b^2=777.51

Taking square root on both sides

\sqrt{b^2}=\sqrt{777.51}\\b=27.88

Rounding off to nearest tenth

b = 27.9 inches

The width of the TV is 27.9 inches

Keywords: Pythagoras theorem, square root

Learn more about square root at:

  • brainly.com/question/2154850
  • brainly.com/question/2367554

#LearnwithBrainly

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Answer:

Square root of the variance of the "number of daily parking tickets"

Step-by-step explanation:

The Standard Deviation is a measure of how spread out numbers are.

Its symbol is σ (the Greek letter sigma)

The formula is easy: it is the square root of the Variance.

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3 years ago
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Answer:

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Step-by-step explanation:

draw a straight line to make it a triangle then you know that angles in a triangle add up to 180.

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180-140=40

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What is the slope of the line that contains the points (-4;2) and (6;-3) ?
tester [92]

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Answer:

1)\ \ 4h^2-13h+6\\2)\ \ 7x^3y^2-x^2y+1\\3)\ \ -7n+2\\4)\ \ -8m+4

Step-by-step explanation:

1.

Simplify the expression by combining like terms. Remember, like terms have the same variable part, to simplify these terms, one performs operations between the coefficients. Please note that a variable with an exponent is not the same as a variable without the exponent. A term with no variable part is referred to as a constant, constants are like terms.

2h^2-7h+2h^2-h+6+4h-9h

(2h^2+2h^2)+(-7h-h+4h-9h)+(6)

4h^2-13h+6

2.

Use a very similar method to solve this problem as used in the first. Please note that all of the rules mentioned in the first problem also apply to this problem; for that matter, the rules mentioned in the first problem generally apply to any pre-algebra problem.

8x^3y^2-7x^2y+8x-4-x^3y^2+2x^2y+4x^2y-8x+5

(8x^3y^2-x^3y^2)+(-7x^2y+2x^2y+4x^2y)+(8x-8x)+(-4+5)

7x^3y^2-x^2y+1

3.

Use the same rules as applied in the first problem. Also, keep the distributive property in mind. In simple terms, the distributive property states the following (a(b+c)=(a)(b)+(a)(c)=ab+ac). Also note, a term raised to an exponent is equal to the term times itself the number of times the exponent indicates. In the event that the term raised to an exponent is a constant, one can simplify it. Apply these properties here,

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4.

The same method used to solve problem (3) can be applied to this problem.

\frac{1}{2}(10-8m+6m^2)-(3m^2+4m-7)-2^3

\frac{1}{2}(10-8m+6m^2)-(3m^2+4m-7)-(2)(2)(2)

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5-4m+3m^2-3m^2-4m+7-8

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