Coeffitiant of 3x-y+4

Answers

Answer 1
Answer:

Answer:

3

Step-by-step explanation:

....,.................

Answer 2
Answer: Your answer Would be A

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Jayden’s new puppy weighed 11 1/4 pounds at 4 weeks old. At 16 weeks old, the puppy weighed 1 2/3 times more.How much did Jayden’s puppy weigh at 16 weeks old?
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2 3/4 divided by 4 1/8 in fraction form

Need Assistance With This
*Please Show Work*​

Answers

Answer:

a =7.5

Step-by-step explanation:

Since this is a right triangle, we can use the Pythagorean theorem

a^2+ b^2 = c^2  where a and b are the legs and c is the hypotenuse

a^2 + 10 ^2 = 12.5^2

a^2 + 100  =156.25

Subtract 100 from each side

a^2 = 56.25

Take the square root of each side

sqrt(a^2) = sqrt( 56.25)

a =7.5

You have 8 quarts of brown stock. You need 3 cups to make one serving of braised short ribs. How many servings can you make? (don't include partial portions) ​

Answers

Answer:

You can make 2 servings.

Step-by-step explanation:

Hi there!

3 cups makes one serving, so 2 servings require 6 cups. Since you don't have 9 cups to make 3 servings, and you don't want partial portions, you can only make 2 servings with 6 cups and have 2 cups left over.

Have a great day!

(I'd also appreicate it if I got a rating and maybe a Thanks please!)

A rectangle with an area of 12 square units, what are the demensions?

Answers

Answer:

area= base(length)*height(width)

possible dimensions

4 *3, 3*4

6*2, 2*6

12*1, 1*12

When we use letters to represent unknown numbers, we call the equation

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A value the replace a number that you can't see
It's called variables, that's what we see

-5+i/2i how do I break this down?​

Answers

Answer:

i2 = -1

Step-by-step explanation:

5i ⋅i⋅(−2i)= −

10 ⋅ i2 ⋅ i= − 10 ⋅ (−1) ⋅ i = 10i

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)