What is the surface area of this figure?

Slav-nsk [51]

The answer would be -144

7 0

2 years ago

A triangle has a base of two to the third power units and a height of chicken fourth power units what is the area of the triangl

kenny6666 [7]

I think you had an auto correct problem

4 0

2 years ago

Find the indicated limit, if it exists.

Paha777 [63]

The limit is the y value that the graph approaches from both sides as x approaches the target limit

it they approach different values from the left and right sides, the limit does not exist

we want to find the limit as x approaches 5

we are given that f(x)=5-x for x<5 and f(5)=8 and f(x)=x+3 for x>5

evaluate them all for f(5)
f(x)=5-x
f(5)=5-5
f(5)=0

f(5)=8

f(x)=x+3
f(5)=5+3
f(5)=8

it approaches 0 and 8 from both sides
the limit doesn't exist
D

8 0

2 years ago

Match the graph given to the right to its function.

KengaRu [80]

SOLUTION

Below is an example of the graph

Hence the answer is Reciprocsl function

6 0

11 months ago

The height (in meters) of a projectile shot vertically upward from a point 4 m above ground level with an initial velocity of 25

Assoli18 [71]

Answer:

(a) The velocity after 2 second is 5.9 m/s

The velocity after 4 second is -13.7 m/s.

(b) The projectile reaches its maximum height after 2.60 s of projection.

(c)The maximum height that is attained by the projectile is 37.18 m.

(d)Therefore the projectile hits the ground after 5.36 seconds of projection.

(e)The velocity of the projectile when it hits the ground is 27.03 m/s

Step-by-step explanation:

Given that, a projectile shot vertically upward from a point 4 m above the ground with a initial velocity of 25.5 m/s.

The height of the projectile after t seconds is

$h=4+25.5t-4.9t^2$

where h is in meter.

(a)

We use the formula

v=u+at

V= final velocity

u = initial velocity = 25.5 m/s

a = acceleration= acceleration due to gravity= 9.8 m/s²

Since the object moves upward direction and acceleration due to gravity is downward direction. So here a= -9.8 m/s.

v(2)= 25.5+(-9.8)×2

=25.5-19.6

=5.9 m/s

And when t= 4

v(4)= 25.5+(-9.8)×4

=25.5-39.2

= -13.7 m/s

The velocity after 2 second is 5.9 m/s

The velocity after 4 second is -13.7 m/s.

(b)

At its maximum height,the velocity of the projectile is zero. i.e v=0

∴0=25.5+(-9.8)t

⇒9.8t=25.5

$\Rightarrow t=\frac{25.5}{9.8}$

⇒t = 2.60 s

The projectile reaches its maximum height 2.60 s after projection.

(c)

To find the maximum height, we are putting t= 2.60 in this equation $h=4+25.5t-4.9t^2$ .

$\therefore h= 4+(25.5\times 2.60)-(4.9\times 2.60^2)$

=37.18 m

The maximum height that is attained by the projectile is 37.18 m.

(d)

When the projectile hits the ground the height will be zero i.e h=0

From the equation of height we get

$\therefore h=0=4+25.5t-4.9t^2$

$\Rightarrow 4+25.5t-4.9t^2=0$

$\Rightarrow t=\frac{-25.5\pm\sqrt{25.5^2-4(-4.9).4}}{2(-4.9)}$

⇒t= -0.15 ,5.36

Therefore it hits the ground after 5.36 seconds of projection.

(e)

To find the velocity we use the formula v=u+at

Here v = final velocity=?

u=25.5 m/s,

t = 5.36 s

a= -9.8m/s²

v=25.5+(-9.8)5.36

= -27.03 m/s

Negative sign denoted that the motion of the projectile is downward direction.

The velocity of the projectile when it hits the ground is 27.03 m/s.

6 0

3 years ago