Please help!!!

X

tresset_1 [31]

Step-by-step explanation:

x and y are roughly in a linear relationship.

when x increases by 2 units, then also y increases by more or less 2 units.

when x increases by 3 units, then also y increases by more or less 3 units.

it is not precise, but a good approximation model.

3 0

3 years ago

Which is the inverse of the function a(d)=5d-3? And use the definition of inverse functions to prove a(d) and a-1(d) are inverse

Drupady [299]

Answer:

$a'(d) = \frac{d}{5} + \frac{3}{5}$

$a(a'(d)) = a'(a(d)) = d$

Step-by-step explanation:

Given

$a(d) = 5d - 3$

Solving (a): Write as inverse function

$a(d) = 5d - 3$

Represent a(d) as y

$y = 5d - 3$

Swap positions of d and y

$d = 5y - 3$

Make y the subject

$5y = d + 3$

$y = \frac{d}{5} + \frac{3}{5}$

Replace y with a'(d)

$a'(d) = \frac{d}{5} + \frac{3}{5}$

Prove that a(d) and a'(d) are inverse functions

$a'(d) = \frac{d}{5} + \frac{3}{5}$ and $a(d) = 5d - 3$

To do this, we prove that:

$a(a'(d)) = a'(a(d)) = d$

Solving for $a(a'(d))$

$a(a'(d)) = a(\frac{d}{5} + \frac{3}{5})$

Substitute $\frac{d}{5} + \frac{3}{5}$ for d in $a(d) = 5d - 3$

$a(a'(d)) = 5(\frac{d}{5} + \frac{3}{5}) - 3$

$a(a'(d)) = \frac{5d}{5} + \frac{15}{5} - 3$

$a(a'(d)) = d + 3 - 3$

$a(a'(d)) = d$

Solving for: $a'(a(d))$

$a'(a(d)) = a'(5d - 3)$

Substitute 5d - 3 for d in $a'(d) = \frac{d}{5} + \frac{3}{5}$

$a'(a(d)) = \frac{5d - 3}{5} + \frac{3}{5}$

Add fractions

$a'(a(d)) = \frac{5d - 3+3}{5}$

$a'(a(d)) = \frac{5d}{5}$

$a'(a(d)) = d$

Hence:

$a(a'(d)) = a'(a(d)) = d$

7 0

3 years ago

I need help with this

Ad libitum [116K]

It’s 60 degrees so the first one

8 0

3 years ago

Read 2 more answers

Greg is designing a clock face. Using the center of the clock face as the origin, he keeps its diameter at 10 units. Match the p

Nat2105 [25]

The center of the clock is taken as the origin.The clock is a circle with a diameter 10 units.Radius is half the diameter .Radius = 10 ÷2= 5 units.

The clock is divided in four quadrants .On x axis y=0 and on y axis x=0.

When it is 12 o'clock the hour hand is on positive of y axis.Coordinates of the point at 12 o'clock=(0,5)

When it is 3 o 'clock the hour hand is on positive of x axis .Coordinate of the point at 3o'clock is (5,0)

When it is 6 o'clock the hour hand is on negative of y axis .The coordinates of the point at 6o'clock is (0,-5)

At 9o'clock the hour hand is on negative of x axis .The coordinate of the point at 6o'clock is(-5,0)

3 0

4 years ago

Read 2 more answers

-9y - 2 (2 - 7y) + 4

poizon [28]

Answer:

-9y-2(2-7y)+4 expand the brackets using-2

Step-by-step explanation:

-9y-4+14y+4

arrange the like terms

-9y+14y-4+4

=5y becoz -4+4=0

ans=5y

7 0

3 years ago

Read 2 more answers