What scale factor was applied to the first rectangle to get the resulting image?

A box is launched upward from the ground at a speed of 57 feet per second. Calculate the height in feet of the box at 3 seconds.

expeople1 [14]

The vertical height traveled is given by
h = ut +- (1/2)g*t^2
where
u = 57 ft/s, initial speed
t = tme, s
g = 32 ft/s^2

After 3 seconds,
h = (57 ft/s)*(3 s) - (1/2)*(32 ft/s^2)*(3 s)^2
= 171 - 144
= 27 ft

Answer: 27 ft

6 0

3 years ago

A car’s speed before a sudden stop can be estimated by the formula

Veseljchak [2.6K]

Answer:

S = 30 mph

Step-by-step explanation:

A car’s speed before a sudden stop can be estimated by the formula :

$S=\sqrt{30gk}$

Here,

S = speed of the car in mph,

g = drag factor, and

k = length of skid mark in feet

If drag factor is 1, k = 30 feet

So,

$S=\sqrt{30\times 1\times 30}\\\\S=30\ mph$

Therefore, the correct option is (a) 30 mph.

6 0

3 years ago

In Youngtown, the population reached 4420 people in 1993. In 1998, there were 5600 residents. What was the population in the yea

Ede4ka [16]

Growth rate = (Present - Past)/Past Plugging in what we know Growth rate =(5600â’4420)/4420 Thus Growth rate=.26697 Now we can plug the growth rate into our first formula which gives us P=5600e^(.26697â‹…2) Solve for P and we get P = 9551.58 however since you can not have .58 of a person we round down to 9551. So Youngtown will have 9551 citizens in the year 2000

6 0

3 years ago

The front and back covers of a text book are each 0.3 thick between the cover are 200 sheets of paper each 0.008 cm thick how hi

shepuryov [24]

Answer: 1,000

Step-by-step explanation:

The height of 10 stacks of books would be a 1,000 because if it's 0.3 thick between covers of 200 sheets then 10 books would be a 1,000.

5 0

3 years ago

MATH HELP PLEASE!!! PLEASE HELP!!!

grandymaker [24]

The answer is: [C]: " f(c) = $\frac{9}{5}$ c + 32 " .
________________________________________________________

Explanation:
________________________________________________________
Given the original function:

" c(y) = (5/9) (x − 32) " ; in which "x = f" ; and "y = c(f) " ;
________________________________________________________
→ Write the original function as: " y = (5/9) (x − 32) " ;

Now, change the "y" to an "x" ; and the "x" to a "y"; and rewrite; as follows:
________________________________________________________
x = (5/9) (y − 32) ;

Now, rewrite THIS equation; by solving for "y" ; in terms of "x" ;
_____________________________________________________
→ That is, solve this equation for "y" ; with "c" as an "isolated variable" on the
"left-hand side" of the equation:

We have:

→ x = " ( $\frac{5}{9}$ ) * (y − 32) " ;

Let us simplify the "right-hand side" of the equation:
_____________________________________________________

Note the "distributive property" of multiplication:
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a(b + c) = ab + ac ; AND:

a(b – c) = ab – ac .
__________________________________________

As such:
__________________________________________

" ( $\frac{5}{9}$ ) * (y − 32) " ;

= [ ( $\frac{5}{9}$ ) * y ] − [ ( $\frac{5}{9}$ ) * (32) ] ;

= [ ( $\frac{5}{9}$ ) y ] − [ ( $\frac{5}{9}$ ) * ( $\frac{32}{1})$ " ;

= [ ( $\frac{5}{9}$ ) y ] − [ ( $\frac{(5*32)}{(9*1)}$ ] ;

= [ ( $\frac{5}{9}$ ) y ] − [ ( $\frac{(160)}{(9)}$ ] ;

= [ ( $\frac{5y}{9}$ ) ] − [ ( $\frac{(160)}{(9)}$ ] ;

= [ $\frac{(5y-160)}{9}$ ] ;
_______________________________________________
And rewrite as:

→ " x = $\frac{(5y-160)}{9}$ " ;

We want to rewrite this; solving for "y"; with "y" isolated as a "single variable" on the "left-hand side" of the equation ;

We have:

→ " x = $\frac{(5y-160)}{9}$ " ;

↔ " $\frac{(5y-160)}{9}$ = x ;

Multiply both sides of the equation by "9" ;

9 * $\frac{(5y-160)}{9}$ = x * 9 ;

to get:

→ 5y − 160 = 9x ;

Now, add "160" to each side of the equation; as follows:
_______________________________________________________

→ 5y − 160 + 160 = 9x + 160 ;

to get:

→ 5y = 9x + 160 ;

Now, divided Each side of the equation by "5" ;
to isolate "y" on one side of the equation; & to solve for "y" ;

→ 5y / 5 = (9y + 160) / 5 ;

to get:

→ y = (9/5)x + (160/5) ;

→ y = (9/5)x + 32 ;

→ Now, remember we had substituted: "y" for "c(f)" ;

Now that we have the "equation for the inverse" ;
→ which is: " (9/5)x + 32" ;

Remember that for the original ("non-inverse" equation); "y" was used in place of "c(f)" . We have the "inverse equation"; so we can denote this "inverse function" ; that is, the "inverse" of "c(f)" as: "f(c)" .

Note that "x = c" ;
_____________________________________________________
So, the inverse function is: " f(c) = (9/5) c + 32 " .
_____________________________________________________

The answer is: " f(c) = $\frac{9}{5}$ c + 32 " ;
_____________________________________________________
→ which is:

→ Answer choice: [C]: " f(c) = $\frac{9}{5}$ c + 32 " .
_____________________________________________________

6 0

3 years ago