All categories

2 years ago

Together, a necklace and bracelet cost $192. Find the price of each if the necklace costs 3 times as much as the bracelet

1 answer:

5 0

Ok set the necklace as X and the bracelet as Y
X+Y=192
X=3Y
Plug in 3Y for X in the first equation
4y=192
Y=48$
Plug that into your second equation
X=3(48)
X=144$

You might be interested in

A swimming pool is being filled with a hose. The hose fills the pool at a constant rate. The pool's water level rises 3 inches i

Marizza181 [45]

0.3 inches per hour, if I did the math correctly.

5 0

3 years ago

tom was using wire of the following thicknesses .33mm , .275mm, .25mm, .3mm for some electrical work. order the wire from thicke

Lena [83]

Thickest to thinnest....
0.33m , 0.3mm, 0.275mm, 0.25mm

5 0

3 years ago

Two cylinders have equal diameters. are their volumes also equal? is the same true for cones and spheres?

Paraphin [41]

Although two cylinders may have equal diameters, their volumes are not necessarily equal, as the volume of a cylinder is dependent also on the cylinder's height. The same holds true for cones. Spheres, however, are, by nature, proportional. Therefore, if two spheres have the same diameter, they also have the same volumes.

5 0

2 years ago

Read 2 more answers

60 = 9x - 21 What is the solution to this equation plz hurry

madam [21]

Answer:

x=9

Step-by-step explanation:

hope this helps i think it's right 60=9x - 21 you add 21 to both sides and get 81 = 9x then divide 81 by 9 and get x=9

8 0

2 years ago

Read 2 more answers

A ship leaves port at 10 miles per hour, with a heading of N 35° W. There is a warning buoy located 5 miles directly north of th

Leno4ka [110]

The value of the angle subtended by the distance of the buoy from the

port is given by sine and cosine rule.

The bearing of the buoy from the is approximately 307.35°

Reasons:

Location from which the ship sails = Port

The speed of the ship = 10 mph

Direction of the ship = N35°W

Location of the warning buoy = 5 miles north of the port

Required: The bearing of the warning buoy from the ship after 7.5 hours.

Solution:

The distance travelled by the ship = 7.5 hours × 10 mph = 75 miles

By cosine rule, we have;

a² = b² + c² - 2·b·c·cos(A)

Where;

a = The distance between the ship and the buoy

b = The distance between the ship and the port = 75 miles

c = The distance between the buoy and the port = 5 miles

Angle ∠A = The angle between the ship and the buoy = The bearing of the ship = 35°

Which gives;

a = √(75² + 5² - 2 × 75 × 5 × cos(35°))

By sine rule, we have;

$\displaystyle \frac{a}{sin(A)} = \mathbf{ \frac{b}{sin(B)}}$

Therefore;

$\displaystyle sin(B)= \frac{b \cdot sin(A)}{a}$

Which gives;

$\displaystyle sin(B) = \mathbf{\frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }}$

$\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx 37.32^{\circ}$

Similarly, we can get;

$\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx \mathbf{ 142.68^{\circ}}$

The angle subtended by the distance of the buoy from the port, C is therefore;

C ≈ 180° - 142.68° - 35° ≈ 2.32°

By alternate interior angles, we have;

The bearing of the warning buoy as seen from the ship is therefore;

Bearing of buoy ≈ 270° + 35° + 2.32° ≈ 307.35°

Learn more about bearing in mathematics here:

brainly.com/question/23427938

5 0

2 years ago