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Stels [109]
3 years ago
12

Suppose (1,19) is on the graph of y = f (x). Which of the following points lies on the graph of the transformed function y = f(1

/5x)?

Mathematics
1 answer:
o-na [289]3 years ago
8 0

Answer:

(5,19) lies on the graph of the transformed function y = f(1/5x)

Step-by-step explanation:

Suppose (1,19) is on the graph of y = f (x)

the graph of the transformed function y = f(1/5x)

y=f(\frac{1}{5}x)

1/5 is multiplied with x  in f(x)

1/5 is less than 1 so there will be a horizontal stretch in the graph by the factor of 1/5

To make horizontal stretch we change the point

f(x)=f(bx) then (x,y) --->( x/b,y)

We divide the x coordinate by the fraction 1/5

(1,19) ----> (\frac{1}{\frac{1}{5}} , 19)= (5,19)

So (5,19) lies on the graph of the transformed function y = f(1/5x)

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A direct variation function contains the points( -8,-6) and (12,9). Which equation represents the function ?
natulia [17]

Answer:

y = (3/4)x

Step-by-step explanation:

1)  Find the slope of the line.  As we go from ( -8,-6) to (12,9), x increases by 20 and y increases by 15.  Thus, the slope, m, is m = rise / run = 15/20 = 3/4

2) Recognize that the y-intercept is zero (0) because this is direct variation; the line goes thru the origin.

3) write the equation of the line:  y = mx + b becomes y = (3/4)x

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3 years ago
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Convert 30 mph to meters per s3cond and round to the nearest tenth​
Tamiku [17]

Answer:

13.4

Step-by-step explanation:

Divide the number of meters in a mile (1,609.344) by the number of seconds in an hour (3,600): 1,609.344 divided by 3,600 equals 0.44704. One mile per hour equals 0.44704 meters per second.

8 0
3 years ago
25 POINTS AND BRAINLIEST PLEASE HELP ASAP
dexar [7]
M is a midpoint of BC so:

M=\left(\dfrac{\boxed{2}\boxed{a}+a}{\boxed{2}},\dfrac{\boxed{0}+b}{2}\right)=\left(\dfrac{\boxed{3}\boxed{a}}{\boxed{2}},\dfrac{\boxed{b}}{\boxed{2}}\right)

Length of MA:

MA=\sqrt{\left(\dfrac{\boxed{3}a}{2}\boxed{-}\boxed{0}\right)^2+\left(\dfrac{\boxed{b}}{2}\boxed{-}\boxed{0}\right)^2}=\\\\\\=
\sqrt{\left(\dfrac{\boxed{3}a}{\boxed{2}}\right)^2+\left(\dfrac{b}{2}\right)^2}=\sqrt{\dfrac{\boxed{9}a^2}{\boxed{4}}+\dfrac{\boxed{b}^2}{\boxed{4}}}

Length of NB:

NB=\sqrt{\left(\dfrac{a}{2}\boxed{-}\boxed{2}a\right)^2+\left(\dfrac{b}{2}\boxed{-}\boxed{0}\right)^2}=\\\\\\=\sqrt{\left(\dfrac{a}{2}\boxed{-}\dfrac{\boxed{4}\boxed{a}}{2}\right)^2+\left(\dfrac{b}{2}-\boxed{0}\right)^2}=\\\\\\
\sqrt{\left(\dfrac{-3a}{2}\right)^2+\left(\dfrac{b}{\boxed{2}}\right)^2}=\sqrt{\dfrac{\boxed{9}a^2}{\boxed{4}}+\dfrac{\boxed{b}^2}{\boxed{4}}}
5 0
3 years ago
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Please someone what is the total surface area of this pryamid ​
Dimas [21]

Answer:

\large\boxed{C.\ (25+25\sqrt3)\ in^2}

Step-by-step explanation:

We have the square and four equilateral triangles.

The formula of an area of a squre:

A_S=a^2

a - length of side

The formula of an area of an equilateral triangle:

A_T=\dfrac{a^2\sqrt3}{4}

a - length of side

Clculate the areas:

SQURE:

A_S=x^2\ in^2

TRIANGLE:

A_T=\dfrac{x^2\sqrt3}{4}\ in^2

The SURFACE AREA of a square pyramid:

S.A.=A_S+4A_T\\\\S.A.=x^2+4\cdot\dfrac{x^2\sqrt3}{4}=x^2+x^2\sqrt3

Put x = 5:

S.A.=5^2+5^2\sqrt3=(25+25\sqrt3)\ in^2

7 0
3 years ago
The formula for the volume of a cone is V = r²h, where r is the radius of the
pshichka [43]

Answer:

<u>h = V/r²</u>

Step-by-step explanation:

The formula is :

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Solving :

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  • V/r² = r²h/r²
  • <u>h = V/r²</u>
3 0
2 years ago
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