MATHEMATICS COLLEGE

Collect like terms
−
8
d
2
−
3
d
2

Answers

Answer 1
Answer:

Answer:

8,2,2,3

d,d,

HOPE THIS HELPS
Answer 2
Answer:

Answer:

5d²

Step-by-step explanation:

Given

8d² - 3d² ← collect like terms by subtracting the coefficients, that is

(8 - 3)d²

= 5d²

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John ordered five large size pizzas for a party. Five kids each ate 1/10 slice, and twelve adults each at 3/10 slices.a) Find the total number of slices consumed.
41/10
b) Find the leftover pizza if any.

Answers

Answer:

a) 4 1/10 slices eaten

b) cannot be solved because we don't know how many slices in each pizza

Step-by-step explanation:

5*1/10=5/10

12*3/10=36/10

36/10+5/10=41/10

41/10=4 1/10 slices

Please and thank you this is a must so plz answer!

Answers

Answer:

The one on the top right

Step-by-step explanation:

For 1 ticket its $10. That graph shows each number of tickets by 10.

2. 4x - 2y = -6
-6x + 2y = 2

Answers

Answer:

x = 2, y =7

Step-by-step explanation:

4x - 2y = -6

-6x + 2y = 2

Add the equations together

4x - 2y = -6

-6x + 2y = 2

-----------------------

-2x = -4

Divide each side by -2

-2x/-2 = -4/-2

x = 2

now find y

-6x+2y =2

-6(2) +2y =2

-12+2y =2

Add 12 to each side

-12+12+2y = 2+12

2y =14

Divide by 2

2y/2 =14/2

y =7

Subtract the following military time: 0545 - 0058

Answers

You subtract it and get 487

Answer:

i think it would be 0447

Step-by-step explanation:

A large tank is filled to capacity with 600 gallons of pure water. Brine containing 5 pounds of salt per gallon is pumped into the tank at a rate of 6 gal/min. The well-mixed solution is pumped out at a rate of 12 gallons/min. Find the number A(t) of pounds of salt in the tank at time t. A(t)

Answers

Salt flows into the tank at a rate of

(5 lb/gal) * (6 gal/min) = 30 lb/min

The volume of solution in the tank after t min is

600 gal + (6 gal/min - 12 gal/min)*(t min) = 600 - 6t gal

which means salt flows out at a rate of

(A(t)/(600 - 6t) lb/gal) * (12 gal/min) = 2 A(t)/(100 - t) lb/min

Then the net rate of change of the salt content is modeled by the linear differential equation,

A'(t)=30-(2A(t))/(100-t)

Solve for A:

A'+(2A)/(100-t)=30

Multiply both sides by the integrating factor, \frac1{(100-t)^2}:

(A')/((100-t)^2)+(2A)/((100-t)^3)=(30)/((100-t)^2)

\left(\frac A{(100-t)^2}\right)'=(30)/((100-t)^2)

Integrate both sides:

\frac A{(100-t)^2}=(30)/(100-t)+C

\implies A(t)=30(100-t)+C(100-t)^2

The tank starts with no salt, so A(0) = 0 lb. This means

0=30(100)+C(100)^2\implies C=-\frac3{10}

and the particular solution to the ODE is

A(t)=30(100-t)-\frac3{10}(100-t)^2=\frac3{10}t(100-t)

Enter the range of values for x

Answers

Greetings from Brasil...

See the attached figure. The smaller the θ angle, the smaller the AB side will be. If the angle θ = 90º, then AB = 25. As θ < 90, then AB < 25

5X - 10 < 25

5X < 25 + 10

X < 35/5

X < 7

The AB side can be neither zero nor negative. So

5X - 10 > 0

5X > 10

X > 10/5

X > 2

2 < X < 7
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