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ryzh [129]
4 years ago
14

Help!!!!! stuck on this

Mathematics
1 answer:
8090 [49]4 years ago
7 0
I hope this helps you




Pythagorean



hypothenus ^2=base^2+height ^2



hypothenus ^2=15^2+20^2



hypothenus ^2=225+400



hypothenus ^2=625



hypothenus ^2=25
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1^-100

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An athlete wants to supplement his daily diet with beef and plant proteins. The content of iron, calcium, and vitamins (in milli
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Total spoon required is mentioned in the Attachment.

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Let a and b be real numbers where a b 0. Which of the following functions could represent the graph below?
MAXImum [283]

The polynomial function that could represent the graph is f(x) = x²(x - a)(x - b)² with four turning points

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

The maximum number of turning points of a polynomial function is always one less than the degree of the function.

Hence the polynomial represented by the graph is of degree 5 since it has 4 turning points. The polynomial function that could represent the graph is f(x) = x²(x - a)(x - b)²

Find out more on equation at: brainly.com/question/2972832

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8 0
2 years ago
Marta drew a rectangular poster.
lilavasa [31]

Answer:

1/2 yard

Step-by-step explanation:

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7 0
3 years ago
Given that f.x 3x-2 over x+1 g[x] x +5 evaluate f[-4] and gf [-2]
Jobisdone [24]

The value of f[ -4 ] and g°f[-2] are \frac{14}{3} and 13 respectively.

<h3>What is the value of f[-4] and g°f[-2]?</h3>

Given the function;

  • f(x) = \frac{3x-2}{x+1}
  • g(x)=x+5
  • f[ -4 ] = ?
  • g°f[ -2 ] = ?

For f[ -4 ], we substitute -4 for every variable x in the function.

f(x) = \frac{3x-2}{x+1}\\\\f(-4) = \frac{3(-4)-2}{(-4)+1}\\\\f(-4) = \frac{-12-2}{-4+1}\\\\f(-4) = \frac{-14}{-3}\\\\f(-4) = \frac{14}{3}

For g°f[-2]

g°f[-2] is expressed as g(f(-2))

g(\frac{3x-2}{x+1}) =  (\frac{3x-2}{x+1}) + 5\\\\g(\frac{3x-2}{x+1}) =  \frac{3x-2}{x+1} + \frac{5(x+1)}{x+1}\\\\g(\frac{3x-2}{x+1}) =  \frac{3x-2+5(x+1)}{x+1}\\\\g(\frac{3x-2}{x+1}) =  \frac{8x+3}{x+1}\\\\We\ substitute \ in \ [-2] \\\\g(\frac{3x-2}{x+1}) =  \frac{8(-2)+3}{(-2)+1}\\\\g(\frac{3x-2}{x+1}) =  \frac{-16+3}{-2+1}\\\\g(\frac{3x-2}{x+1}) =  \frac{-13}{-1}\\\\g(\frac{3x-2}{x+1}) =  13

Therefore, the value of f[ -4 ] and g°f[-2] are \frac{14}{3} and 13 respectively.

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6 0
2 years ago
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