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2 years ago

A ticket company charges a 4% service fee on all orders. How much is the service fee for a ticket that costs $65?

Mathematics

1 answer:

solong [7]2 years ago

6 0

Answer:

$16.25

Step-by-step explanation:

4 goes into 65 16 r1

which = 16.25

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Lines m and n are parallel and m<1=140*<br> What is m<7?

Arada [10]

$m\angle 7=180-m\angle 5\\
m\angle 5=m\angle 1=140\\
m\angle 7=180-140\\
m\angle 7=40$

6 0

3 years ago

Calculate the simple interest earned on an investment of $26,000 at 8% p.a. Over 3 years.

Annette [7]

I=prt
I=26,000×0.08×3
I=6,240

6 0

3 years ago

Prove that<br><img src="https://tex.z-dn.net/?f=%20%7B2%7D%5E%7B3%7D%20%20%5Ctimes%20%20%7B3%7D%5E%7B0%7D%20%20%2B%20%20%7B3%7D%

KonstantinChe [14]

Answer:

Step-by-step explanation:

Anything raise to the power zero is 1

Therefore,

$LHS\\\\ =2^3 \times 3^0 + 3^2 \times 2^0\\\\8 \times 1 + 9 \times 1 \\\\17$

LHS = RHS

Hence Proved

5 0

3 years ago

Suppose Albers Elementary School has 32 teachers and Bothell Elementary School has 61 teachers. If the total number of teachers

Contact [7]

Answer: 31 teachers

Step-by-step explanation: let the number of teachers that teach in both school be x

By using venn diagram:

32 - x + x + 61 - x = 62

93 - x = 62

X = 93 - 62

X = 31

7 0

4 years 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