All categories

3 years ago

A compact-disc case, or jewel case is 14 cm long, 12.5 cm wide, and 1 cm tall. What is the volume of the jewel case?

Mathematics

1 answer:

puteri [66]3 years ago

4 0

V = 14 × 12.5 × 1 => V = 175 $cm^{3}$ is the volume of the jewel case.

You might be interested in

Someone help!!! PLEASE BE QUICK.... <br> **BONUS POINTS**

Tatiana [17]

Answer:

884cm^2

Step-by-step explanation:

top triangle area=21*24/2=252

central triangle area=16*24/2=192

left n right triangle area (each)=22*20/2=220

total area of wood needed=252+192+220+220

=884cm^2

6 0

2 years ago

Read 2 more answers

I need help. Give an explanation, please!

AURORKA [14]

Answer:

y=35+x+15x/100. since x is the money earn so 15percent of x

3 0

2 years ago

Read 2 more answers

What is the value of x? 35x – 13 x = x – 1

Papessa [141]

35x – 13 x = x – 1

⇒ 22 x = x - 1

⇒ 22x - x = -1

⇒ 21x = -1

⇒ x = -1/21

Thus,the value of x is -1/21

6 0

3 years ago

Read 2 more answers

Solve the quadratic equation numerically (using tables of x- and y- values).

cricket20 [7]

Answer:

x= -3 or x= -4

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

In ΔABC, if m ∠A = m∠C, m∠B = ß (where ß is an acute angle), and BC = x, which expression gives the length of b, the side opposi

chubhunter [2.5K]

Answer:

The length of b is $\sqrt{2x^2(1-cos\beta)}$ .

Step by step explanation:

Given information: In ΔABC, ∠A =∠C, ∠B = ß (where ß is an acute angle), and BC = x.

Since two angles are same therefore triangle ABC is an isosceles triangle and side AB and BC are congruent.

$AB=BC=x$

According to Law of cosine

$b^2=a^2+c^2-2ac\cos B$

$b^2=x^2+x^2-2(x)(x)\cos \beta$

$b^2=2x^2-2x^2\cos \beta$

$b^2=2x^2(1-\cos \beta)$

$\sqrt{2x^2(1-cos\beta)}$

Therefore the length of b is $\sqrt{2x^2(1-cos\beta)}$ .

6 0

2 years ago