Ask question

Ask question

All categories

3 years ago

9

My watch is 5 minutes slow, but i think it is 3 minutes fast. I arrive 'on time' according to my calculations to catch the 1:15

pm train What is the real time when I arrive at the station?

1 answer:

faltersainse [42]3 years ago

4 0

1:15 pm + 5 minutes =1:20 PM You add 5 minutes because your watch is slow then subtract 3 minutes because you thought the watch was 3 minutes fast. real time is 1:17 PM
1:20 pm - 3 minutes = 1:17 PM

You might be interested in

the area of a square floor on a scale drawing is 100 square centimeters, and the scale of the drawing is 1 cm: 2 ft. what is the

kondaur [170]

The square root of 100 is 10. The drawing must be 10 cm by 10 cm.

1 cm = 2 ft, so 10 cm must equal 20 feet.

20 * 20 = 400 square feet :)

3 0

3 years ago

Solve for x: 5x - 2 = 3x + 8

gayaneshka [121]

5x-2=3x+8

X=5

That's your answer

8 0

3 years ago

Read 2 more answers

Will give brainiest!!!

azamat

Answer:

i take bite :3

Step-by-step explanation:

6 0

3 years ago

Determine what shape is formed for the given coordinates for ABCD, and then find the perimeter and area as an exact value and ro

Helga [31]

Answer:

Part 1) The shape is a trapezoid

Part 2) The perimeter is $25(4+\sqrt{2})\ units$ or approximately $135.4\ units$

Part 3) The area is $937.5\ units^2$

Step-by-step explanation:

step 1

Plot the figure to better understand the problem

we have

A(-28,2),B(-21,-22),C(27,-8),D(-4,9)

using a graphing tool

The shape is a trapezoid

see the attached figure

step 2

Find the perimeter

we know that

The perimeter of the trapezoid is equal to

$P=AB+BC+CD+AD$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Find the distance AB

we have

A(-28,2),B(-21,-22)

substitute in the formula

$d=\sqrt{(-22-2)^{2}+(-21+28)^{2}}$

$d=\sqrt{(-24)^{2}+(7)^{2}}$

$d=\sqrt{625}$

$d_A_B=25\ units$

Find the distance BC

we have

B(-21,-22),C(27,-8)

substitute in the formula

$d=\sqrt{(-8+22)^{2}+(27+21)^{2}}$

$d=\sqrt{(14)^{2}+(48)^{2}}$

$d=\sqrt{2,500}$

$d_B_C=50\ units$

Find the distance CD

we have

C(27,-8),D(-4,9)

substitute in the formula

$d=\sqrt{(9+8)^{2}+(-4-27)^{2}}$

$d=\sqrt{(17)^{2}+(-31)^{2}}$

$d=\sqrt{1,250}$

$d_C_D=25\sqrt{2}\ units$

Find the distance AD

we have

A(-28,2),D(-4,9)

substitute in the formula

$d=\sqrt{(9-2)^{2}+(-4+28)^{2}}$

$d=\sqrt{(7)^{2}+(24)^{2}}$

$d=\sqrt{625}$

$d_A_D=25\ units$

Find the perimeter

$P=25+50+25\sqrt{2}+25$

$P=(100+25\sqrt{2})\ units$

simplify

$P=25(4+\sqrt{2})\ units$ ----> exact value

$P=135.4\ units$

therefore

The perimeter is $25(4+\sqrt{2})\ units$ or approximately $135.4\ units$

step 3

Find the area

The area of trapezoid is equal to

$A=\frac{1}{2}[BC+AD]AB$

substitute the given values

$A=\frac{1}{2}[50+25]25=937.5\ units^2$

4 0

3 years ago

The distance by road from town A to town B is 257 km. What is 50% of that distance?

Marina86 [1]

Answer: 128.5km

Step-by-step explanation:

Since we are given the information that the distance by road from town A to town B is 257 km. To get 50% of the distance, we simply have to multiply the distance given by 50%. This will be:

= 50% × 257km

= 50/100 × 257km

= 0.5 × 257km

= 128.5km

Therefore, 50% of the distance is 128.5km.

3 0

3 years ago