alex saved $16 this month. if he decides to donate 3/4 of his savings to charity. hiw much money will alex be donating?

Two bicyclists are 7/8 of the way through a mile long tunnel when a train approaches the closer end at 40 mph. The riders take o

Anastaziya [24]

Answer:

the cyclists rode at 35 mph

Step-by-step explanation:

Assuming that the cyclists stopped, and accelerated instantaneously at the same speed than before but in opposite direction , then

distance= speed*time

since the cyclists and the train reaches the end of the tunnel at the same time and denoting L as the length of the tunnel :

time = distance covered by cyclists / speed of cyclists = distance covered by train / speed of the train

thus denoting v as the speed of the cyclists :

7/8*L / v = L / 40 mph

v = 7/8 * 40 mph = 35 mph

v= 35 mph

thus the cyclists rode at 35 mph

6 0

3 years ago

On the first part of her trip Natalie rode her bike 16 miles and on the second part of the trip she rode her bike 42 miles. Her

-BARSIC- [3]

Answer:

14 mph ( average speed during the second part of the trip )

Step-by-step explanation:

Let´s call "x" the average speed during the first part then

t = 5 hours

t = t₁ + t₂ t₁ and t₂ times during part 1 and 2 respectively

l = t*v ( distance is speed by time ) t = l/v

First part

t₁ = 16/x and t₂ = 42 / ( x + 6)

Then

t = 5 = 16/x + 42 /(x + 6)

5 = [ 16 * ( x + 6 ) + 42 * x ] / x* ( x + 6 )

5 *x * ( x + 6 ) = 16*x + 96 + 42 x

5*x² + 30*x - 58*x - 96 = 0

5*x² - 28*x - 96 = 0

We obtained a second degree equation, we will solve for x and dismiss any negative root since negative time has not sense

x₁,₂ = [28 ± √ (28)² + 1920 ] / 10

x₁,₂ = ( 28 ± √2704 )/ 10

x₁ = 28 - 52 /10 we dismiss that root

x₂ = 80/10

x₂ = 8 mph average speed during the first part, and the average speed in the second part was 6 more miles than in the firsst part. then the average spedd dring the scond part was 8 + 6 = 14 mph

6 0

3 years ago

Harold has 4 classes each morning. Each class is 1 hour long, and there are 10 minutes between classes. The first class is at 8

Viktor [21]

The fourth class ends at 12:30 pm

Solution:

Given that Harold has 4 classes each morning

Each class is 1 hour long, and there are 10 minutes between classes

The first class is at 8 A.M

To find: Time at which fourth class ends

Since each is 1 hour long and 10 minutes gap between classes

First class = 8 A.M to 9 A.M

Second class = 9:10 A.M to 10 : 10 AM

Third class = 10 : 20 AM to 11 : 20 AM

Fourth class = 11 : 30 AM to 12 : 30 PM

Thus the fourth class ends at 12:30 pm

4 0

2 years ago

an office supply company sells two types of printers. they charge $95 for one of the printers and $125 for the other. if the com

Ivenika [448]

Answer:

22 of the 95$ ones and 10 of the 125

Step-by-step explanation:

22 times 95 = 2090

10 times 125 = 1250

2090+1250=3340

hope this helped

4 0

2 years ago

HELP ASAP TIMED !!!! What is the exact value of cos(c + d), given Sine c = StartFraction 24 Over 25 EndFraction for c in Quadran

Keith_Richards [23]

9514 1404 393

Answer:

D

Step-by-step explanation:

In the second quadrant, ...

c = 180° -arcsin(24/25) ≈ 106.26°

In the third quadrant, ...

d = 360° -arccos(-3/4) ≈ 221.41°

Then cos(c+d) = cos(327.67°) ≈ 0.84498

This is a positive irrational number, greater than 21/100, so the only reasonable choice is the last one:

$\dfrac{21+24\sqrt{7}}{100}\approx 0.84498$

_____

Perhaps you want to work this out using the trig identities.

cos(c) = -√(1 -sin(c)²) = -7/25

sin(d) = -√(1 -cos(d)²) = -(√7)/4

Then the desired cosine is ...

cos(c+d) = cos(c)cos(d) -sin(c)sin(d)

cos(c+d) = (-7/25)(-3/4) -(24/25)(-√7/4)

cos(c+d) = (21 +24√7)/100 . . . . matches choice D

8 0

2 years ago