5. A soccer ball is kicked from the ground level with an upward velocity of 60 feet per second. The

A bee flies at 20 feet per second directly to a flower bed from its hive. The bee stays at the flower bed for 15 minutes, and th

oee [108]

Answer:

a. Speed of flight * ( total spent away from hive - time stayed on flower bed)

b. The flower bed is 2,400 ft from the hive

Step-by-step explanation:

a. Mathematically, the distance will be ;

the speed * time taken

Given that he stays a total of 17 minutes away from the hive and he stayed 15 minutes in the flower bed, the time it used on the flower bed will be 17 minutes - 15 minutes = 2 minutes

So the distance from the flower bed to the hive is;

Speed of flight * ( total spent away from hive - time stayed on flower bed)

b. We want to find the distance of the flower bed from the hive

That will be;

20 ft per second * 2 minutes (120 seconds)

= 20 * 120 = 2,400 ft

4 0

3 years ago

Graph ARST with vertices R(6, 6), S(3, -6), and T(0, 3) and its image after a

padilas [110]

Answer:

The answer is the second figure and the vertices of Δ R'S'T' are:

R' = (-6 , 6) , S' = (-3 , -6) , T' = (0 , 3)

Step-by-step explanation:

* Lets revise some transformation

- If point (x , y) reflected across the x-axis

∴ Its image is (x , -y)

- If point (x , y) reflected across the y-axis

∴ Its image is (-x , y)

- If point (x , y) reflected across the line y = x

∴ Its image is (y , x)

- If point (x , y) reflected across the line y = -x

∴ Its image is (-y , -x)

- Now we can solve the problem

∵ R = (6 , 6) , S = (3 , -6) , T = (0 , 3), they are the vertices of ΔRST

- The triangle RST is reflected over the y-axis

- According to the rule above the signs of x-coordinates will change

∵ R = (6 , 6)

∴ Its image is (-6 , 6)

∵ S = (3 , -6)

∴ Its image is (-3 , -6)

∵ T = (0 , 3)

∴ Its image is (0 , 3)

* Now lets look to the figure to find the correct answers

- The image of Δ RST is ΔR'S'T'

∵ The vertices of the image of ΔRST are:

R' = (-6 , 6) , S' = (-3 , -6) , T' = (0 , 3)

* The answer is the second figure

7 0

3 years ago

Show all work and reasoning

Natalija [7]

Split up the interval [2, 5] into $n$ equally spaced subintervals, then consider the value of $f(x)$ at the right endpoint of each subinterval.

The length of the interval is $5-2=3$ , so the length of each subinterval would be $\dfrac3n$ . This means the first rectangle's height would be taken to be $x^2$ when $x=2+\dfrac3n$ , so that the height is $\left(2+\dfrac3n\right)^2$ , and its base would have length $\dfrac{3k}n$ . So the area under $x^2$ over the first subinterval is $\left(2+\dfrac3n\right)^2\dfrac3n$ .

Continuing in this fashion, the area under $x^2$ over the $k$ th subinterval is approximated by $\left(2+\dfrac{3k}n\right)^2\dfrac{3k}n$ , and so the Riemann approximation to the definite integral is

$\displaystyle\sum_{k=1}^n\left(2+\frac{3k}n\right)^2\frac{3k}n$

and its value is given exactly by taking $n\to\infty$ . So the answer is D (and the value of the integral is exactly 39).