Can someone help me here

Mark had 3 quarts of milk how many pints of milk did he have

mamaluj [8]

Mark had 6 pints of milk.

7 0

3 years ago

Provide the reflected ordered pair for each of the given ordered pairs. Reflect the ordered pair ( − 4 , 5 ) (-4, 5) across the

levacccp [35]

Answer:

a) (4,5)

b) (0,-3)

Step-by-step explanation:

We have to perform the following reflection over given ordered pair.

a) Reflect the ordered pair (-4,5) across the y-axis

Reflection over y-axis:

$r: (x,y)\rightarrow (-x,y)$

Thus, (-4,5) will be reflected over y-axis as

$r: (-4,5)\rightarrow (-(-4),5) = (4.5)$

b) Reflect the ordered pair (0,3) across the y-axis

Reflection over x-axis:

$r: (x,y)\rightarrow (x,-y)$

Thus, (0,3) will be reflected over x-axis as

$r: (0,3)\rightarrow (0,-(3)) = (0,-3)$

6 0

3 years ago

Write an equation of a line parallel to line EF below in slope-intercept form that passes through the point (2, 6).

N76 [4]

Answer:

y = -⅔x + ²²/₃

Step-by-step explanation:

1. Calculate the slope of EF

Assume E is at (-2, 5) and F is at (1, 3).

m₁ = (y₂ - y₁)/(x₂ - x₁) = (3 - 5)/(1 -(-2)) = -⅔

2. Calculate the slope of the parallel line

m₂ = m₁ = -⅔

3. Calculate its y-intercept

y = mx + b

6 =-⅔(2) + b

Simplify

6 = -⁴/₃ + b

Add ⁴/₃ to each side

b = 6 + ⁴/₃ = ¹⁸/₃ + ⁴/₃ = ²²/₃

The y-intercept is at (0,²²/₃).

4. Write the equation for the line

y = -⅔x + ²²/₃

The diagram shows EF in red. The parallel line in black passes through (2, 6) with a y-intercept at (0,²²/₃).

7 0

3 years ago

A tank contains 200 liters of fluid in which 50 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu

mel-nik [20]

At the start, the tank contains A(0) = 50 g of salt.

Salt flows in at a rate of

(1 g/L) * (5 L/min) = 5 g/min

and flows out at a rate of

(A(t)/200 g/L) * (5 L/min) = A(t)/40 g/min

so that the amount of salt in the tank at time t changes according to

A'(t) = 5 - A(t)/40

Solve the ODE for A(t):

A'(t) + A(t)/40 = 5

e^(t/40) A'(t) + e^(t/40)/40 A(t) = 5e^(t/40)

(e^(t/40) A(t))' = 5e^(t/40)

e^(t/40) A(t) = 200e^(t/40) + C

A(t) = 200 + Ce^(-t/40)

Given that A(0) = 50, we find

50 = 200 + C ==> C = -150

so that the amount of salt in the tank at time t is

A(t) = 200 - 150 e^(-t/40)

7 0

3 years ago

The length of fencing required for a rectangular garden in terms of its width, x, if it’s length is 2 units more than 1.5 times

JulijaS [17]

Answer:

$5x+4$ units.

Step-by-step explanation:

Let x be the width of rectangle.

We have been given that the length of garden is 2 units more than 1.5 times it’s width. So length of the rectangle will be: $1.5x+2$ .

To find the length of total fencing we need to figure out perimeter of rectangle with width x and length $1.5x+2$ .

Since we know that perimeter of a rectangle is two times the sum of its length and width.

$\text{Perimeter of rectangle}=2(\text{Length+Width})$

Upon substituting length and width of garden in above formula we will get,

$\text{Perimeter of rectangular garden}=2(1.5x+2+x)$

$\text{Perimeter of rectangular garden}=2(2.5x+2)$

Upon using distributive property we will get,

$\text{Perimeter of rectangular garden}=2*2.5x+2*2$

$\text{Perimeter of rectangular garden}=5x+4$

Therefore, the length of required fencing will be $5x+4$ units.

4 0

3 years ago