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3 years ago

Greg run 5 miles in 38 minutes. At the same rate, how many miles would he run in 57 minutes?

Mathematics

1 answer:

makkiz [27]3 years ago

8 0

Answer:

ince you are just starting, i would say anywhere from 40-45 minutes, you should be averaging about 7-8 minutes per mile.

Step-by-step explanation:

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If a little girl has 43 water bottles and 4 friends how much do each person get

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Each person can equally get 10 water bottles. With a remainder of 3. It might also be a fraction

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3 years ago

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Joy is going to buy food for a class reunion. there are 25 guests and each guest will eat 250 grams of chicken in the table .How

ikadub [295]

Answer:

6250

Step-by-step explanation:

25×250

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3 years ago

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Determine whether the graph of each pair of equations are parrallel, perpendicular, or neither

Luba_88 [7]

Answer: 1: parrallel

2:perpendicular

3; neither

4; perpendicular. 5: parrallel

Step-by-step explanation:

7 0

3 years ago

Pls see attached question

zimovet [89]

Answer:

Step-by-step explanation:

put it on a coorinate grid,

circle circumference = 2*pi*r = 100.53

so the arc 19.2/100.53 = about 0.15 times the circumference of the circle

0.15*2pi = angle AOB in radians = 0.97

length DE is like 7.44

angle DOE is approx 0.485 radians beacuse it's half of angle AOB

area of shaded region is 804.2477 probably. I'm not sure about this last one sorry I think my calculator messed up

6 0

3 years ago

Solve using law of sines or law of cosines!

malfutka [58]

Answer:

Part 5) The length of the ski lift is $1.15\ miles$

Part 6) The height of the tree is 18.12 m

Step-by-step explanation:

Part 5)

Let

A -----> Beginning of the ski lift

B -----> Top of the mountain

C -----> Base of mountain

we have

$b=0.75\ miles$

$A=20\°$

$C=180\°-50\°=130\°$ ----> by supplementary angles

Find the measure of angle B

Remember that the sum of the interior angles must be equal to 180 degrees

$B=180\°-A-C$

substitute

$B=180\°-20\°-130\°=30\°$

Applying the law of sines

$\frac{b}{sin(B)}=\frac{c}{sin(C)}$

substitute

$\frac{0.75}{sin(30\°)}=\frac{c}{sin(130\°)}$

$c=\frac{0.75}{sin(30\°)}(sin(130\°))$

$c=1.15\ miles$

Par 6)

see the attached figure with letters to better understand the problem

Applying the law of sines in the right triangle BDC

In the right triangle BDC 20 degrees is the complement of 70 degrees

$\frac{BC}{sin(70\°)}=\frac{x}{sin(20\°)}$

$BC=(sin(70\°))\frac{x}{sin(20\°)}$ -----> equation A

Applying the law of sines in the right triangle ABC

In the right triangle ABC 50 degrees is the complement of 40 degrees

$\frac{BC}{sin(40\°)}=\frac{x+15}{sin(50\°)}$

$BC=(sin(40\°))\frac{x+15}{sin(50\°)}$ -----> equation B

Equate equation A and equation B and solve for x

$(sin(70\°))\frac{x}{sin(20\°)}=(sin(40\°))\frac{x+15}{sin(50\°)}\\\\2.7475x=0.8391(x+15)\\\\2.7475x=0.8391x+12.5865\\\\2.7475x-0.8391x=12.5865\\\\x=6.60\ m$

Find the value of BC

$BC=(sin(70\°))\frac{6.6}{sin(20\°)}$

$BC=18.12\ m$

therefore

The height of the tree is 18.12 m

5 0

4 years ago