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Aloiza [94]
3 years ago
6

What transformations to the linear parent function, f(x) = x, give the function

Mathematics
1 answer:
77julia77 [94]3 years ago
7 0

Answer:

A. Shift down 2 units.

B. Vertically stretch by a factor of 4.

Step-by-step explanation:

Given the function

f(x)=x

If we stretch y vertically by a factor of m, we have: y=m·f (x)

Therefore:

Vertically stretching f(x) by a factor of 4, we have: 4x.

Next, if we take down f(x) by k units we have: y= f(x)-k

Therefore: Taking down 4x by 2 units, we obtain:

g(x)=4x-2

Therefore, Options A and B applies.

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Find the distance between these points.<br> R(-1,0), S(8,6)<br> V(26)<br> V(85)<br> 3V(13)
Thepotemich [5.8K]

Answer:

The distance is equal to 3\sqrt{13}\ units

Step-by-step explanation:

we know that

the formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

we have

R(-1,0)\\S(8,6)  

substitute the values

d=\sqrt{(6-0)^{2}+(8+1)^{2}}

d=\sqrt{(6)^{2}+(9)^{2}}

d=\sqrt{36+81}

d=\sqrt{117}\ units

Simplify

d=3\sqrt{13}\ units

4 0
3 years ago
17. Admission prices to Cinema I to see a movie are $9.50 for an adult and $6.50 for a child. The admission charge at Cinema II
maksim [4K]

Answer:

a. 9.5x + 6.5(x+c) < 8   when c>0

b. Must be one child more than the no. of adults.

Step-by-step explanation:

For Cinema 1:

for adult = $9.50

for child = $6.50

For Cinema 2:

Per person regardless of age = $8.00

First of all, we will find out the condition when per person rates in both cinema are equal.

Assume x = no. of adults

y = no. of children

Rate per person in Cinema I = Rate per person in Cinema II

(9.5x + 6.5y)/(x+y)   =   8

9.5x + 6.5y = 8(x+y)

9.5x + 6.5y = 8x + 8y

9.5x-8x = 8y-6.5y

=> x = y

So rates are equal when no. of adults equals no. of children

For Cinema I to have better rates, no. of children should be atleast 1 more than the no. of adult. In this way the rate per person of Cinema I will be less than 8

Hence we form an inequality when y = x+c and c > 0

9.5x + 6.5(x+c) < 8   when c>0

Hence there must be 1 more children than the no. of adults attending Cinema I for it to be a better deal.

4 0
3 years ago
Read 2 more answers
Solve the proportion.
melisa1 [442]

Answer:

you use older of operations to solve this

3 0
3 years ago
Read 2 more answers
Joseph and Hanna are 250 feet apart when they start walking toward one another. They are walking at the same speed, so whenever
Evgen [1.6K]
I dont know the answer
4 0
4 years ago
Ann has 30% less money than Mary, and Jessica has 70% more than Mary. If the 3 of them have a (added up) total of $102, how much
Roman55 [17]
Let x be the amount of Ann money, y be Jessica's money and z Mary money. 
30% of z equates 0.3*z. 
Ann has 30% less than z so: x = z - 0.3z = 0.7z.
70% of z equates 0.7z. 
"Jessica has 70% more than Mary " we deduce the equation: y =z+0.7z= 1.7z.
Signe the total is $102, we deduce the equation x+y+z = 102. 
x=0.7z and y=1.7z then 0.7z+1.7z+z=102, therefore 3.4z=102 then 
z =102/3.4 = $30.
x=0.7*30= 21 and y=1.7*30=51.

Ann has $21, Mary has $30 and Jessica has $51.
  
3 0
3 years ago
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