All categories

3 years ago

Justine enrolls in a music program. The cost of the proram includes a one time registration fee of $45 and a fee of $9.

Mathematics

2 answers:

Anton [14]3 years ago

3 0

Im not sure but i think its 45+9 x X

small x= times
big X = scenario

icang [17]3 years ago

3 0

Justin must pay a one time $45 fee and each class is $9 so 45+9x

You might be interested in

The seventh-grade class is building target areas for a PE activity. The bases for the game will be in a circular shape. The diam

lys-0071 [83]

By definition, the area of a circle is given by:
$A = \pi * r ^ 2
$
Where,
r: radius of the circle.
Substituting the values in the given expression, we have:
$A = 3.14 * (3) ^ 2

 A = 28.26 feet ^ 2$
Rounding to the nearest tenth we have:
$A=28.3feet^2$
Answer:
Approximately needed:
$A=28.3feet^2$

4 0

3 years ago

Read 2 more answers

What is100*87 hurry plz

shtirl [24]

The answer is 8700. there you go sir or ma'm!

6 0

2 years ago

Read 2 more answers

Mrs. Williams is knitting a blanket for her newborn granddaughter. The blanket is 2.25 meters long and 1.8 meters wide. What is

LuckyWell [14K]

To find the area, you will need to times the length with the width.

So,

2.25*1.8=4.05 meters

1m=100cm

4.05m=405cm

Hope this helps.

5 0

2 years ago

A variable is at the _______ level of measurement if it allows for the values of the variable to be arranged in a specific orde

nignag [31]

The answer would be <span>ordianl.

Hope this helps !

Photon</span>

4 0

3 years ago

From the side view, a gymnastics mat forms a right triangle with other angles measuring 60° and 30°. The gymnastics mat extends

valkas [14]

Answer: The mat is 4.33 ft high off the ground.

Step-by-step explanation:

Since we have given that

Angle of elevation with the first triangle = 30°

Angle of elevation with the second triangle = 60°

Length at which gymnastics mat extends across the floor = 5 feet

so, As shown in the figure:

We need to find the height of the mat off the ground.

If CD = 5 ft,

Let, AB = y, DC = x.

In Δ ABC,

$\tan 60^\circ=\frac{AB}{BC}\\\\\frac{\sqrt{3}}{2}=\frac{y}{x}\\\\x=\frac{y}{\sqrt{3}}$

Similarly, in Δ ACD,

$\tan 30^\circ=\frac{AB}{BD}\\\\\frac{1}{\sqrt{3}}=\frac{y}{x+5}\\\\\frac{1}{\sqrt{3}}=\frac{y}{\frac{y}{\sqrt{3}}+5}\\\\\frac{1}{\sqrt{3}}=\frac{y\sqrt{3}}{y+5\sqrt{3}}\\\\3y=y+5\sqrt{3}\\\\2y=5\sqrt{3}\\\\y=\frac{5\sqrt{3}}{2}\\\\y=4.33\ ft$

Hence, the mat is 4.33 ft high off the ground.

5 0

3 years ago

Read 2 more answers