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lara31 [8.8K]
3 years ago
14

Kevin is 2 times as old as Gabriela. 12 years ago, Kevin was 6 times as old as Gabriela. How old is Gabriela now?

Mathematics
1 answer:
aleksley [76]3 years ago
3 0

Answer:

gabrielas age is 15

Step-by-step explanation:

Let Gabriela's age be x, then Kelvins age is 2x

12 years ago, Gabriela's age was x - 12 and Kelvin's age was 6(x - 12)

i.e. 6x - 72 = 2x - 12

4x = 60

x = 60/4 = 15

Therefore, Kelvin's age is 2(15) = 30 years old

please thank me and rate

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Alex wants to fence in an area for a dog park. He has plotted three sides of the fenced area at the points E (1, 5), F (3, 5), a
hram777 [196]

Answer:

(1,1)

Step-by-step explanation:

Given: E, F, G, H denote the three coordinates of the area fenced

To find: coordinates of point H

Solution:

According to distance formula,

length of side joining points (x_1,y_1)\,,\,(x_2,y_2) is equal to \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

So,

EF=\sqrt{(3-1)^2+(5-5)^2}=2\,\,units\\FG=\sqrt{(6-3)^2+(1-5)^2}=\sqrt{9+16}=5\,\,units\\GH=\sqrt{(x-6)^2+(y-1)^2}\\EH=\sqrt{(x-1)^2+(y-5)^2}

Perimeter of a figure is the length of its outline.

EF+FG+GH+EH=16\\2+5+\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16-2-5\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9

Put (x,y)=(1,1)

\sqrt{(1-6)^2+(1-1)^2}+\sqrt{(1-1)^2+(1-5)^2}=9\\\sqrt{25}+\sqrt{16}=9\\5+4=9\\9=9

This is true.

So, the point (1,1) satisfies the equation \sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9

So, point H is (1,1).

7 0
3 years ago
Find an equation of the line perpendicular to the graph of 15x-5y=7 that passes through the point at (0,-4)
shusha [124]
Perpendicular lines have slopes that multiply to -1
get into y=mx+b form
minus 15x both sides and divide by -5
y=3x-7/5
slope is 3

3 times what=-1?
what=-1/3

so the slope is -1/3


so
y=-1/3x+b
we use the point (0,-4)
x=0 and y=-4
-4=-1/3(0)+b
-4=b

y=(-1/3)x-4 is da equation
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3 years ago
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The graph of a sinusoidal function intersects its midline at (0,5) and then has a maximum point at (\pi,6)
Natalija [7]
First, let's use the given information to determine the function's amplitude, midline, and period. 

Then, we should determine whether to use a sine or a cosine function, based on the point where x=0.

Finally, we should determine the parameters of the function's formula by considering all the above.
     
                      Determining the amplitude, midline, and period 

The midline intersection is at y=5 so this is the midline. 

The maximum point is 1 unit above the midline, so the amplitude is 1. 

The maximum point is π units to the right of the midline intersection, so the period is 4 * π.
 
                            Determining the type of function to use 

Since the graph intersects its midline at x=0, we should use thesine function and not the cosine function. 

This means there's no horizontal shift, so the function is of the form -

a sin(bx)+d

Since the midline intersection at x=0 is followed by a maximumpoint, we know that a > 0.

The amplitude is 1, so |a| = 1. Since a >0 we can conclude that a=1. 

The midline is y=5, so d=5. 

The period is 4π so b = 2π / 4π = 1/2 simplified. 

f(x)1 sin   (\dfrac{1}{3}x) +5  = Solution
8 0
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