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3 years ago

(3 2/3 + 17 1/5) - ( 11 3/5 + 2 1/3) =

Mathematics

2 answers:

ELEN [110]3 years ago

6 0

6 14/15 take the common factor of each and solve

Ipatiy [6.2K]3 years ago

4 0

Answer:

6 14/15

Step-by-step explanation:

I would say simplifie the paratensies.

First:

3 2/3+ 17 1/5 =

Common factor of 3 and 5 is 15

3 10/15 + 17 3/15 = 20 13/15

Second:

11 3/5 + 2 1/3 =

Common factor of 3 and 5 is 15

11 9/15 + 2 5/15 = 13 14/15

20 13/15 - 13 14/15 = 7 13/15 - 14/15 = 7 - 1/15 = 6 14/15

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If cos(x) = Three-fourths and tan(x) < 0, what is cos(2x)?

makvit [3.9K]

Step-by-step explanation:

The value of sin(2x) is \sin(2x) = - \frac{\sqrt{15}}{8}sin(2x)=−

How to determine the value of sin(2x)

The cosine ratio is given as:

\cos(x) = -\frac 14cos(x)=−

Calculate sine(x) using the following identity equation

\sin^2(x) + \cos^2(x) = 1sin

(x)+cos

(x)=1

So we have:

\sin^2(x) + (1/4)^2 = 1sin

(x)+(1/4)

\sin^2(x) + 1/16= 1sin

(x)+1/16=1

Subtract 1/16 from both sides

\sin^2(x) = 15/16sin

(x)=15/16

Take the square root of both sides

\sin(x) = \pm \sqrt{15/16

Given that

tan(x) < 0

It means that:

sin(x) < 0

So, we have:

\sin(x) = -\sqrt{15/16

Simplify

\sin(x) = \sqrt{15}/4sin(x)=

sin(2x) is then calculated as:

\sin(2x) = 2\sin(x)\cos(x)sin(2x)=2sin(x)cos(x)

So, we have:

\sin(2x) = -2 * \frac{\sqrt{15}}{4} * \frac 14sin(2x)=−2∗

∗

This gives

\sin(2x) = - \frac{\sqrt{15}}{8}sin(2x)=−

6 0

2 years ago

Read 2 more answers

3=1 entonces cuanto es 1+1+1

natka813 [3]

Step-by-step explanation:

si 3=1

1+1+1 = 3+3+3

3=9

4 0

3 years ago

Can someone please help me with these questions!! Thanks,

fiasKO [112]

6a. 1 - 2sin(x)² - 2cos(x)² = 1 - 2(sin(x)² +cos(x)²) = 1 - 2·1 = -1

6c. tan(x) + sin(x)/cos(x) = tan(x) + tan(x) = 2tan(x)

6e. 3sin(x) + tan(x)cos(x) = 3sin(x) + (sin(x)/cos(x))cos(x) = 3sin(x) +sin(x) = 4sin(x)

6g. 1 - cos(x)²tan(x)² = 1 - cos(x)²·(sin(x)²)/cos(x)²) = 1 -sin(x)² = cos(x)²

8 0

3 years ago

HELP PLEASE!! i need help on A and B. PLEASE DONT ANSWER IF YOU DONT KNOW!!

alexandr1967 [171]

A.
( 2.5 , 11 )
( 5 , 11 )
( 4.5 , 8 )
( 2 , 8 )
( 7 , 11 )
( 7 , 8 )
( 12 , 11 )
( 4.5 , 6 )
( 4.5 , 2 )
( 2 , 2 )
( 2 , 6 )
( 8 , 2 )
( 8 , 6 )
( 11 , 1 )
( 11 , 7 )

4 0

3 years ago

A tank initially has 300 gallons of a solution that contains 50 lb. of dissolved salt. A brine solution with a concentration of

belka [17]

Let s(t) be the amount of salt in the tank at time t. Then s(0) = 50 lb.

Salt flows into the tank at a rate of

(2 gal/min) (6 lb/gal) = 12 lb/min

and flows out at a rate of

(2 gal/min) (s(t)/300 lb/gal) = s(t)/150 lb/min

so that the net rate at which salt is exchanged through the tank is

ds(t)/dt = 12 - s(t)/150 … (lb/min)

Solve for s(t). The DE is separable, so we have

ds/dt = 12 - s/150

150 ds/dt = 1800 - s

150/(1800 - s) ds = dt

Integrate both sides to get

-150 ln|1800 - s| = t + C

Solve for s :

ln|1800 - s| = -t/150 + C

1800 - s = exp(-t/150 + C )

1800 - s = C exp(-t/150)

s = 1800 - C exp(-t/150)

Now given that s(0) = 50, we solve for C :

50 = 1800 - C exp(-0/150)

50 = 1800 - C

C = 1750

Then the amount of salt in the tank at any time t ≥ 0 is

s(t) = 1800 - 1750 exp(-t/150)

To find the time it takes for the tank to hold 100 lb of salt, solve for t in

100 = 1800 - 1750 exp(-t/150)

1700 = 1750 exp(-t/150)

34/35 = exp(-t/150)

ln(34/35) = -t/150

t = -150 ln(34/35) ≈ 4.348

So it would take approximately 4.348 minutes.

By the way, we didn't have to solve for s(t), we could have instead stopped with

-150 ln|1800 - s| = t + C

Solve for C - this C is not the same as the one we found using the other method. s(0) = 50, so

-150 ln|1800 - 50| = 0 + C

C = -150 ln|1750|

==> t = 150 ln(1750) - 150 ln|1800 - s|

Then s(t) = 100 lb when

t = 150 ln(1750) - 150 ln(1700)

t = 150 ln(1750/1700)

t = 150 ln(35/34)

t ≈ 4.348

5 0

3 years ago