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oee [108]
2 years ago
5

The subtotal for a TV was $950.00. What does that mean

Mathematics
1 answer:
Elis [28]2 years ago
8 0

The subtotal for a TV was $950.00. so it is the cost of the TV before taxes and other fees are added to it.

<h3>What is subtotal?</h3>

Subtotal is defined as the actual cost of any product when there will not be any taxes or charges are levied on the product.

The subtotal for a TV was $950.00. so it is the cost of the TV before taxes and other fees are added to it.

To know more about Subtotal follow

brainly.com/question/27139464

#SPJ1

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A charity is collecting donations as a part of their yearly fundraising event. They are going door-to-door, asking for a donatio
tino4ka555 [31]

Step-by-step explanation:

well first try by addding the numbers and then dived it by 2

8 0
4 years ago
Carlo buys $14.40 worth of grapefruit. Each grapefruit costs 0.80. (PLz help and yeett) with steps plz
Zina [86]

Answer:

a) 18 grapefruits

b) 6 grapefruits

Step-by-step explanation:

a) n = 14.4 / 0.80 = 18 grapefruits

b) n = (14.4 / 3) / 0.8 = 6 grapefruits

4 0
3 years ago
A red car is driving along the road in the direction of the police car and is 130 feet up the road from the location of the poli
pishuonlain [190]

Complete question:

A police car is located 50 feet to the side of a straight road. A red car is driving along the road in the direction of the police car and is 130 feet up the road from the location of the police car. The police radar reads that the distance between the police car and the red car is decreasing at a rate of 75 feet per second. How fast is the red car actually traveling along the road.

Answer:

The red car is traveling along the road at 80.356 ft/s

Step-by-step explanation:

Given

Police car is 50 feet side off the road

Red car is 130 feet up the road

Distance between them is decreasing at the rate of 75 feet per sec

Let x be how far the police is off the road.

Let y be how far the red car is up the road.  

Let h be the distance between the police and the red car.

This forms a right triangle so we can use the Pythagorean theorem, to solve for h

h² = x² + y²

h² = 50² + 130²

h² = 19400

h = √19400

h = 139.284 ft

Again;

Let dx/dt be how fast the police is traveling toward the road.

Let dy/dt  be how fast the red car is traveling along the road.

Let dh/dt be how fast the distance between the police and the car is decreasing.

Recall that, h² = x² + y² (now differentiate with respect to time, t)

2h(dh/dt) = 2x(dx/dt) + 2y(dy/dt)

divide through by 2

h(dh/dt) = x(dx/dt) + y(dy/dt)

since the police car is not, dx/dt = 0

and dy/dt is the how fast is the red car actually traveling along the road

139.284(75) = 50(0) + 130(dy/dt)

10446.3 = 0 + 130(dy/dt)

dy/dt = 10446.3 / 130

dy/dt = 80.356 ft/s

Therefore, the red car is traveling along the road at 80.356 ft/s

6 0
4 years ago
9 1/2 divided by 3 7/10
OleMash [197]

Solution:

<u>Convert the fractions into improper fractions and solve.</u>

  • 9 1/2 ÷ 3 7/10
  • => 19/2 ÷ 37/10
  • => 19/2 x 10/37
  • => 19 x 5/37
  • => 95/37

<u>Convert the fraction into mixed fraction (Not required):</u>

  • => 95/37 = 37/37 + 37/37 + 21/37
  • => 1 + 1 + 21/37
  • => 2 + 21/37
  • => 2 21/37

The solution to the problem is 2 21/37.

8 0
2 years ago
Read 2 more answers
7. Write an equation for the line that is parallel to the given line and that passes through the given point.
Vlad1618 [11]

Answer:

C. y = 1/2x - 14

Step-by-step explanation:

Find the slope of the original line and use the point-slope formula  y − y 1 = m ( x − x 1 )  to find the line parallel to  y = 1 /2 x − 8 .

y = 1 /2 x − 14                              PARALLEL

...................................................................................................................................................

Find the negative reciprocal of the slope of the original line and use the point-slope formula  y − y 1 = m ( x − x 1 )  to find the line perpendicular to  y = 1 /2 x − 8 .

y = − 2 x − 29                               PERPENDICULAR

8 0
2 years ago
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