4a+6b=10
4a=10-6b
a=(10-6b)/4
a=(5-3b)/2

2a-4b=12
2((5-3b)/2)-4b=12
(5-3b)-4b=12
5-7b=12
-7b=7
b=-1

2a-4b=12
2a-4(-1)=12
2a+4=12
2a=8
a=4

a=4, b=-1 

12a=48

7 0

3 years ago

Read 2 more answers

enter an equation in point-slope form for the line. Slope is 7 and (-2,-1) is on the line. The equation of the line in point slo

lutik1710 [3]

Answer:

y + 1 = 7(x + 2)

General Formulas and Concepts:

Algebra I

Point-Slope Form: y - y₁ = m(x - x₁)

x₁ - x coordinate
y₁ - y coordinate
m - slope

Step-by-step explanation:

Step 1: Define

Slope m = 7

Point (-2, -1)

Step 2: Write Equation

Substitute in variables into general form.

y + 1 = 7(x + 2)

7 0

3 years ago

If p(a) = 0.60 and p(b) = 0.20, then a and b are independent events if

sp2606 [1]

Answer: B. P(A and B) =0.12

Step-by-step explanation:

7 0

1 year ago

A tank with a capacity of 100 gallons initially contains 50 gallons of water with 10 pounds of salt in solution. fresh water ent

steposvetlana [31]

Let $A(t)$ be the amount of salt (in pounds) in the tank at time $t$ . We're given that $A(0)=10\text{ lb}$ .

The rate at which the amount of salt in the tank changes is given by the ODE

$A'(t)=\dfrac{2\text{ gal}}{1\text{ min}}\cdot\dfrac{0\text{ lb}}{1\text{ gal}}-\dfrac{1\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{50+(2-1)t\text{ gal}}$

$A'(t)+\dfrac{A(t)}{50+t}=0$

$(50+t)A'(t)+A(t)=0$

$\bigg((50+t)A(t)\bigg)'=0$

$(50+t)A(t)=C$

$A(t)=\dfrac C{50+t}$

Given that $A(0)=10$ , we find that

$10=\dfrac C{50+0}\implies C=500$

so that the amount of salt in the tank is described by

$A(t)=\dfrac{500}{50+t}$

The tank will be filled when $50+t=100$ , or after $t=50$ minutes. At this time, the amount of salt in the tank is

$A(50)=\dfrac{500}{50+50}=5\text{ lb}$

7 0

3 years ago