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4 years ago

5x=6+2x....? help please

Mathematics

2 answers:

kenny6666 [7]4 years ago

8 0

2
subtract 2x on both sides u get 3x=6 , divide both sides by 3 , x=2

Andre45 [30]4 years ago

3 0

Step-by-step explanation:

5x=6+2x

transfer 2x to right side so the sign changes

5x-2x=6

3x=6

x=6/3

x=2

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2b) A farmer is building a pen inside a barn. The pen will be in the shape of a right triangle.

kenny6666 [7]

Answer:

Step-by-step explanation:

the sum of any two sides of a right triangle more than any other side

14+15>29, x is less than 29

14+x>15

x>1

15+x>14

x>-1

1<x<29

lets use Pythagorean Theorem

14^(2)+15^(2)=c^2

196+225=c^2

421=c^2

plusminus sqrt(421)=c

distance can't be negative, so:

c=sqrt(421)

c=20.5182845287

sqrt(421) is the largest possible right triangle side

1<x $\leq$ sqrt(421)

B) as we can see from above, the largest possible length of the third side is 14/sqrt(421) or about 20.5182845287ft

C)sqrt(421) by 14 by 15

tan(x1)=15/14

x1=tan^-1(15/14)

x1=47 degrees

tan(x2)=14/15

x2=tan^-1(14/15)

x2=43 degrees

7 0

3 years ago

At time t hours after taking the cough suppressant hydrocodone bitartrate, the amount, A, in mg, remaining in the body is given

Nostrana [21]

Answer:

a. The initial amount was 10 mg.

b. The percentage of the drug leaving the body each hour is 0.18, this is 18% per hour.

c. The amount of drug that remains in the body 6 hours after dosing is 3.04 mg.

d. The time until only 1 mg of the drug remains in the body is 11.6 hours.

Step-by-step explanation:

You know that at time t hours after taking the cough suppressant hydrocodone bitartrate, the amount, A, in mg, remaining in the body is given by :

$A=10*(0.82)^{t}$

a. The initial quantity occurs when time t is the initial t, that is, t is equal to 0. Then:

$A=10*(0.82)^{t}=10*(0.82)^{0}=10*1\\$

A=10

The initial amount was 10 mg.

b. Considering that an exponential growth is determined by:

$A=A0*(1-r)^{t}$ , where A is the amount after a certain number of a certain time, Ao is the initial amount, r is the rate and t is the time, so the percentage of the drug that leaves the body each hour is :

1-r=0.82

Solving:

1-r -1= 0.82 -1

-r= -0.18

r= 0.18

The percentage of the drug leaving the body each hour is 0.18, this is 18% per hour.

c. The amount of drug that remains in the body 6 hours after dosing is when t = 6:

$A=10*(0.82)^{6}$

Solving:

A= 3.04 mg

The amount of drug that remains in the body 6 hours after dosing is 3.04 mg.

d. The time that passes until only 1 mg of the drug remains in the body is calculated taking into account that A = 1 mg:

$1=10*(0.82)^{t}$

Solving:

$0.1=(0.82)^{t}$

㏒ 0.1= t*㏒ 0.82

㏒ 0.1 ÷ ㏒ 0.82= t

11.6 hours= t

The time until only 1 mg of the drug remains in the body is 11.6 hours.

4 0

3 years ago

2(-3y-1)+(2y+7) plz help

MrRissso [65]

2(-3y-1)+(2y+7)

Final result :

5 - 4y

Step by step solution :

Step 1 :

Step 2 :

Pulling out like terms :

2.1 Pull out like factors :

-3y - 1 = -1 • (3y + 1)

Equation at the end of step 2 :

-2 • (3y + 1) + (2y + 7)

Step 3 :

Final result :

5 - 4

6 0

4 years ago

I need the answer and how to do it

irga5000 [103]

You turn 1/3 into 2/6. Because you need the bottom number to match. 3x2=6 then you do the same to the top. 1x2=2. So your answer would be 4/6 but reduced it would be 2/3

3 0

3 years ago

Linda drove 715 miles in 13 hours.

Free_Kalibri [48]

Answer:

9 hours

Step-by-step explanation:

you divide the 715 miles by 13 hours then take the quotient of that (55) and divide the 495 miles by 55 which gets you to your answer, 9.

8 0

3 years ago