3. An acute triangle with angles measuring 60 degrees,62°, and 58 degrees,

The distance between Frankfurt and Prague is 432.5 km. Alex left Frankfurt at 10 AM and started traveling to Prague at 60 km/h.

viva [34]

Answer:

He was already 120 km away from Frankfurt so they will meet at 120 + 150 = 370 km away from Frankfurt

Step-by-step explanation:

We are given that the distance between Frankfurt and Prague is 432.5 km.

Distance bw Frankfurt and prague, s(F&P) = 432.5km

Alex left Frankfurt at 10 AM and tavelled at 60 km/hr. At noon i.e 12 PM Michael started traveling from Prague to Frankfurt, which means that Alex travelled 2 hour before Michael started, so lets calculate that distance first.

At 10 AM:

Time travelled by Alex, t = 2hr

Speed, v = 60km/hr

Distance, s(Alex) = v × t = 60 × 2 = 120 km

At noon 12 PM:

Now that Alex has covered 120 km from Frankfurt, he is now 432.5 - 120 = 312.5 km away from Michael at 12 PM.

Distance between Michael and Alex is now 312.5 km

Now Micahel will start moving towards Frankfurt at a speed of 65km/hr

And Alex is also coming towards Prague at a speed of 60km/hr

As they are moving toward eachother their speed will be taken as cumulative.

Speed, v = 65 + 60 = 125

Distance between the two, s = 312.5km

Time t = $\frac{s}{v}$ = $\frac{312.5}{125}$ = 2.5 hr

In these 2.5 hr Alex will cover a distance at 60km/hr = 2.5 * 60 = 150

He was already 120 km away from Frankfurt so they will meet at 120 + 150 = 370 km away from Frankfurt

5 0

3 years ago

Jackie bought a 3-pack of tubes of glue. She used 0.27 ounce of the glue from the 3-pack

brilliants [131]

Answer:

0.54 ounces in each tube

Step-by-step explanation:

Add the amount she used so we can add it in. 1.35+0.27= 1.62. Then divide it by the number of tube and since it was a 3 pack you divide it by 3. 1.62/3=0.54 so there is 0.54 ounces in each tube.

5 0

3 years ago

Read 2 more answers

Find the LCD of Number 21 and 23.

Alina [70]

The LCD is basically the LCM of the denoenator so

to find LCM of exg 6 and 21, find factors and group
6=2 and 3 (one 2 and one 3)
21=3 and 7 (one 3 and one 7)
lcm=2 and 3 and 7 (include 6 and 21 in the number) (one 2 and one 3 and one 7)

LCM≈LCD
so
17.
denomenators are 2 and x^2
2=2
x^2=x times x
lcm=2 times x times x=2x^2

18.
denomenators are 6 and 9
6=2 and 3
9=3 and 3
LCM=2 times 3 times 3=18

19.
denomenators are z and 7z
z=z
7z=7 and z
lcm=7 times z=7z

20.
denomenators are 5b and 7b^3c
5b=5 and b
7b^3c=7 times b times b times b times c
LCM=7 times 5 times b times b times b times c=7b^3c=7cb^3

21. denomenator=5 and x+2
5=5
x+2=x+2
LCM=5 times (x+2)=5(x+2)=5x+10

22. denomenators are ab and b^2c
ab=a time b
b^2c=b times b times c
LCM=a times b times b times c=ab^2c=acb^2

23. demonenators are m+n and m-n

m+n=m+n
m-n=m-n
LCM=(m+n)(m-n)=m^2-n^2

24.denoenators are k and k^2-2

k=k
k^2-2=k^2-2
LCM=k times (k^2-2)=k^3-2k

5 0

4 years ago

Read 2 more answers

Given the points J(4,-7) and L(-2, 13) find the coordinates of point K on line JL such that the ratio of JK to JL 1:4

Anastasy [175]

Answer:

The answer is K(2.5,-2)

Step-by-step explanation: