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2 years ago

PLZ HELP FAST (multiple choice)

Mathematics

2 answers:

Nataly_w [17]2 years ago

8 0

The answer is 0 (my sentence needs to have 20 characters so I’m typing extra)

never [62]2 years ago

6 0

Answer:

C. 0

Step-by-step explanation: I think if i am wrong im sorry

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A school P is 16km due west of a school Q. what is the bearing of Q from P

Irina-Kira [14]

Answer:

16Km due east of school P

Step-by-step explanation:

Given

A school P is 16km due west of a school Q

Thus, we can say that distance PQ = 16 km.

________________________________

now we have to find the bearing of Q from P

As distance is same

distance PQ = distance QP

Thus,

Distance will remain same of 16 km.

For direction,

If Q is west of P, then P will be east of Q $P------------------>Q$

as shown in figure P is west of Q,

now from point P , Q is west P.

Thus,

Bearing of School Q from P is 16Km due east of school P

7 0

3 years ago

HELP ME PLS Write the ratio 3/8 with a numerator of 9.

Anna35 [415]

Answer:

Step-by-step explanation:

3 9

9 times 8

divided by 3 = denominator

9x8÷3=24

7 0

3 years ago

Read 2 more answers

Suppose two angles in a coordinate plane, one measuring 80° and the other measuring 160°, were both rotated 180° about the origi

Snowcat [4.5K]

C because I just took it

5 0

2 years ago

Read 2 more answers

Find each Of change. Round to the nearest whole percent if necessary. State whether the percent Of change is an increase or 1. O

Alenkasestr [34]

Answer:

1. The percentage change is 74% and it is a percentage decrease

2. The percentage change is 15% and it is a percentage decrease

3. The percentage change is 20% and it is a percentage decrease

4. The percentage change is 43% and it is a percentage increase

5. The percentage change is 40% and it is a percentage decrease

6. The percentage change is 39% and it is a percentage decrease

Step-by-step explanation:

Given

$1.\ Original = 5; New = 1.3$

$2.\ Original = 160; New = 136$

$3.\ Original = 15; New = 12$

$4.\ Original = 7; New = 10$

$5.\ Original = $30; New = $18$

$6.\ Original = 9.6; New = 5.9$

Required

Determine the percentage change and state whether it's an increase or decrease

Percentage change is calculated as:

$\%Change = \frac{New - Original}{Original} * 100\%$

$1.\ Original = 5; New = 1.3$

$\%Change = \frac{1.3 - 5}{1.3} * 100\%$

$\%Change = \frac{-3.7}5} * 100\%$

$\%Change = \frac{-3.7 * 100\%}{5}$

$\%Change = \frac{-370\%}{5}$

$\%Change = -74\%$

The percentage change is 74% and it is a percentage decrease

$2.\ Original = 160; New = 136$

$\%Change = \frac{136 - 160}{160} * 100\%$

$\%Change = \frac{-24}{160} * 100\%$

$\%Change = \frac{-24 * 100\%}{160}$

$\%Change = \frac{-2400\%}{160}$

$\%Change = -15\%$

The percentage change is 15% and it is a percentage decrease

$3.\ Original = 15; New = 12$

$\%Change = \frac{12- 15}{15} * 100\%$

$\%Change = \frac{-3}{15} * 100\%$

$\%Change = \frac{-3* 100\%}{15}$

$\%Change = \frac{-300\%}{15}$

$\%Change = -20\%$

The percentage change is 20% and it is a percentage decrease

$4.\ Original = 7; New = 10$

$\%Change = \frac{10- 7}{7} * 100\%$

$\%Change = \frac{3}{7} * 100\%$

$\%Change = \frac{3* 100\%}{7}$

$\%Change = \frac{300\%}{7}$

$\%Change = 43\%$ -- (approximated)

The percentage change is 43% and it is a percentage increase

$5.\ Original = $30; New = $18$

$\%Change = \frac{18- 30}{30} * 100\%$

$\%Change = \frac{-12}{30} * 100\%$

$\%Change = \frac{-12* 100\%}{30}$

$\%Change = \frac{-1200\%}{30}$

$\%Change = -40\%}$

The percentage change is 40% and it is a percentage decrease

$6.\ Original = 9.6; New = 5.9$

$\%Change = \frac{5.9- 9.6}{9.6} * 100\%$

$\%Change = \frac{-3.7}{9.6} * 100\%$

$\%Change = \frac{-3.7* 100\%}{9.6}$

$\%Change = \frac{-370\%}{9.6}$

$\%Change = -39\%$ -- (approximated)

The percentage change is 39% and it is a percentage decrease

5 0

3 years ago

A person jogged 10 times along the perimeter of a rectangular field at the rate of 12 kilometers per hour for 30 minutes. If the

notsponge [240]

Answer:

The area of the rectangular field is 20,000 square meters.

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

Number of times a person jogged along the rectangular field = 10

Speed of the person jogging = 12 km per hour

Time he/she jogged = 30 minutes = 0.5 hours

Length of the field = twice its width

2. Find the area of the field in square meters.

Step A: Let's calculate the distance the person jogged

Let's recall the formula of the speed:

Speed = Distance/Time

Distance = Speed * Time

Replacing with the values we know:

Distance = 12 * 0.5

Distance = 6 kilometers

Step B: Let's calculate the perimeter of the rectangular field

Distance jogged = ¨Perimeter * 10

6 kilometers = Perimeter * 10

Perimeter = 6/10

Perimeter = 0.6 kilometers = 0.6 * 1,000 = 600 meters

Perimeter = 2 * Length + 2 * Width

x = Width of the rectangular field

2x = Length of the rectangular field

600 = 2 * 2x + 2 * x

600 = 4x + 2x

600 = 6x

x = 600/6 = 100 meters ⇒ 2x = 200 meters

Step C: Let's calculate the area of the rectangular field

Let's recall the formula of the area of a rectangle:

A = Length * Width

A = 200 * 100

A = 20,000 m² (square meters)

7 0

3 years ago

PLZ HELP FAST (multiple choice)​

PLZ HELP FAST (multiple choice)