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2 years ago

A 25% decrease followed by a 50% increase

Mathematics

2 answers:

bogdanovich [222]2 years ago

7 0

Greater, plug in 100 for example. A 25% decrease would be 75, and a 50% increase from 75 is 112.5

m_a_m_a [10]2 years ago

6 0

Answer:greater

Step-by-step explanation:

You might be interested in

Which of the following points is a solution of y > Ixl + 5?

Step2247 [10]

Answer:

B. (1,7)

Step-by-step explanation:

We can substitute the x and y values of each coordinate into the inequality and test if they work.

Let's start with A, 5 being y and 0 being x .

$5 > |0|+5\\5> 0+5\\5 > 5$

5 IS NOT greater than 5, they are the exact same, so A is out.

Let's try B, 1 being x and 7 being y.

$7 > |1| + 5\\7 > 1 + 5\\7 > 6$

7 IS greater than 6, so B. (1,7) does work for this inequality!

Let's do C for fun, when 7 is x and 1 is y.

$1 > |7| + 5\\1>7+5\\1>12$

1 IS NOT greater than 12, it is quite less than 12, so C doesn't work.

Therefore B. (1,7) works for the inequality of $y > |x|+5$ .

Hope this helped!

7 0

3 years ago

2. Consider the circle x^2−4x+y^2+10y+13=0. There are two lines tangent to this circle having a slope of 2/3.

likoan [24]

Answer:

a) The coordinates are (0.431, -2.646) and (3.568,-7.354)

b) The tangent lines are

l₁ = (x-0.431)*2/3 - 2.646

l₂ = (x-3.568)2/3 - 7.354

Step-by-step explanation:

First, lets complete squares, by taking for each cordinate the square of a linear expression

0= x²−4x+y²+10y+13 = (x-2)²+ 4 + (y+5)²-25+13 = (x-2)²+(y+5)² - 8

Hence (x-2)² + (y+5)² = 8

Lets put y in function of x. We should obtain 2 functions f and g that represent the circle.

(y+5)² = 8 - (x-2)² = -x² + 4x + 4

$y = ^+_- \sqrt{-x^2+4x+4} -5$

Thus

$f(x) = \sqrt{-x^2+4x+4} - 5$

$g(x) = -\sqrt{-x^2+4x+4} - 5$

Lets find the derivate of each function and the points in which they reach the value 2/3. In those points the tangent line will have a slope of 2/3.

We may just find the values of x which the derivate of f is either 2/3 or -2/3.

$\frac{-x+2}{\sqrt{-x^2+4x+4}} = ^+_-\frac{2}{3} \Leftrightarrow \frac{x^2-4x+4}{-x^2+4x+4} = \frac{4}{9} \Leftrightarrow 9(x^2-4x+4) = 4(-x^2+4x+4) \\\Leftrightarrow 13x^2-52x+20 = 0$