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3 years ago

How does the function f(x) = a ln x compare to the parent function when |a| > 1?

Mathematics

1 answer:

VladimirAG [237]3 years ago

5 0

Answer: The graph is stretched.

Step-by-step explanation:

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Find the perimeter of Triangle ABC with the verticies A(-5,5) B(3,-5) and C(-5,1

finlep [7]

Hello!

Let's calculate a distance between two points using the Pythagorean theorem:

 $d ^ 2_ {AB} = (x_ {B} - x_ {A}) ^ 2 + (y_ {B} - y_ {A}) ^ 2$

Data:

$x_B = 3 

x_A = -5 

y_B = -5 

y_A = 5 

d_ {AB} =?$

Solving: distance from A to B

$d^ 2_ {AB} = (x_ {B} - x_ {A}) ^ 2 + (y_ {B} - y_ {A}) ^ 2 

d^ 2_ {AB} = (3 - (-5)) ^ 2 + (-5 - 5) ^ 2 

d^ 2_ {AB} = (8) ^ 2 + (-10) ^ 2 

d^ 2_ {AB} = 64 + 100 

d^ 2_ {AB} = 164 

d_{AB} = \sqrt {164} 

d_{AB} = \sqrt {2 ^ 2 * 41} 

\boxed {d_ {AB} = 2 \sqrt {41}}$

Solving:

distance from A to C

$d ^ 2_ {AC} = (x_ {C} - x_ {A}) ^ 2 + (y_ {C} - y_ {A}) ^ 2$

Data:

$x_C = -5
 
x_A = -5
 
y_C = 1 

y_A = 5 

d_ {AC} =?$

$d^ 2_ {AC} = (x_ {C} - x_ {A}) ^ 2 + (y_ {C} - y_ {A}) ^ 2 

d^ 2_ {AC} = (-5 - (-5)) ^ 2 + (1 - 5) ^ 2 

d^ 2_ {AC} = (0) ^ 2 + (-4) ^ 2 

d^ 2_ {AC} = 0 + 16 

d^ 2_ {AC} = 16 

d_ {AC} = \sqrt {16} 

\boxed {d_ {AC} = 4}
$

Solving:

distance from B to C

$d^ 2_ {BC} = (x_ {C} - x_ {B}) ^ 2 + (y_ {C} - y_ {B}) ^ 2$

Data:

$x_C = -5 

x_B = 3 

y_C = 1 

y_B = -5 

d_ {BC} =?$

$d^ 2_ {BC} = (x_ {C} - x_ {B}) ^ 2 + (y_ {C} - y_ {B}) ^ 2
 
d^ 2_ {BC} = (-5 - 3) ^ 2 + (1 - (-5)) ^ 2

d^ 2_ {BC} = (-8) ^ 2 + (6) ^ 2 

d^ 2_ {BC} = 64 + 36

d^ 2_ {BC} = 100 

d_ {BC} = \sqrt {100}

\boxed {d_ {BC} = 10}$

Now let's calculate the perimeter of the triangle ABC, knowing that the perimeter is the sum of the sides, then:

$p = d_ {AB} + d_ {AC} + d_ {BC} 

$

$p = 2\sqrt {41} + 4 + 10$

$
\boxed {\boxed {p = 14 + 2 \sqrt {41}}} \end {array}} \qquad \quad \checkmark$

8 0

3 years ago

A hang glider is soaring over a 100-acre area that consists of thick forest and open fields. In the diagram below, the forested

Sonbull [250]

Answer:0.8 :)))

Step-by-step explanation:

8 0

3 years ago

Solve for x:a(a²+b²)x²+b²x-a

m_a_m_a [10]

Answer:

x = a/(a² + b²) or x = -1/a

Step-by-step explanation:

a(a²+ b²)x² + b²x - a =0

Use the quadratic equation formula:

$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} =\dfrac{-b\pm\sqrt{D}}{2a}$

1. Evaluate the discriminant D

D = b² - 4ac = b⁴ - 4a(a² + b²)(-a) = b⁴ + 4a⁴ + 4a²b² = (b² + 2a²)²

2. Solve for x

$\begin{array}{rcl}x & = & \dfrac{-b\pm\sqrt{D}}{2a}\\\\ & = & \dfrac{-b^{2}\pm\sqrt{(b^{2}+2a^{2})^{2}}}{2a(a^{2} + b^{2})}\\\\ & = & \dfrac{-b^{2}\pm(b^{2}+ 2a^{2})}{2a(a^{2} + b^{2})}\\\\x = \dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}\\\\x =\dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\\end{array}$

$\begin{array}{rcl}x = \large \boxed{\mathbf{\dfrac{a}{a^{2} + b^{2}}}}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2b^{2}- 2a^{2}}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2(a^{2}+ b^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\large \boxed{\mathbf{-\dfrac{1}{a}}}\\\\\end{array}$

5 0

3 years ago

One number exceeds another by 44. the sum of the numbers is 24. what are the numbers?

baherus [9]

Let, the numbers = x,y

x+y= 24 1st eqn

x-y = 44 2nd eqn

24-y-y = 44

-2y = 44-24

y = 20/-2 = -10

substitute that in equation 1st x = 24+10 = 34

so, the numbers would be 34 & -10

4 0

3 years ago

Is 0.102 less than 3/7

Bond [772]

it is less than

3/7 ⇒ 0.42857142857

3 0

3 years ago