2 Points

10. Mowing a lawn Joe can mow a lawn in 4 minutes less time than his sister Karen. Together they can

vovangra [49]

About 24 minutes? because together it would take about 12 mins for her and 8 mins for Joe. then she has to do double the work making it 24. That’s an educated guess sooooo.

6 0

3 years ago

What is the slope of the line

Dennis_Churaev [7]

Answer:

Step-by-step explanation:

(0, 2) (2, 0)

(0-2)/(2-0)= -2/2= -1

y - 2 = -(x - 0)

y - 2 = -x + 0

y = -x + 2

4 0

2 years ago

McCoy Brothers manufactures and sells two products, A and Z in the ratio of 4:2. Product A sells for $75; Z sells for $95. Varia

ExtremeBDS [4]

Answer:

3100 units of Product Z

Step-by-step explanation:

The ratio is 4 is to 2, that means:

4 + 2 = 6 parts total

Now,

We have to cover the fixed cost of 418,500 from profits, that is SP - VC

Where SP is Selling Price and VC is Variable Cost

Product A:

SP = 75

VC = 35

Profit = 40

Product B:

SP = 95

VC = 40

Profit = 55

So, we can sell "4x" of product A and "2x" of product B, and create the equation:

$4x(40) + 2x(55) = 418,500$

Now, we solve for x:

$4x(40) + 2x(55) = 418,500\\160x + 110x = 418,500\\270x=418500\\x=1550$

The amount of Product Z was taken as "2x", so the quantity of Product Z would be:

2(1550) = 3100 units of Product Z

4 0

3 years ago

Irum is sitting on the beach, watching the tide go in and out. Irum's distance from the shoreline (in meters) as a function of t

Nesterboy [21]

Answer:

Hi, the Answer is 0.75.

Step-by-step explanation:

it is 0.75 because if you look on the graph, and you calculate the 3/4 slope between the two, 3/4= 0.75

4 0

3 years ago

Find the perimeter AND area of the parallelogram ABCD. Show all work

stiks02 [169]

Answer-

The perimeter and area of the parallelogram are 19.74 units and 15 sq. units respectively.

Solution-

The co-ordinates of the vertices are,

A = (-2, 3)

B = (4, 0)

C = (1, -1)

D = (-5, 2)

E = (-3, 1)

We can get the side length of the parallelogram by calculating the respective distances by applying distance formula,

$\overline{CD}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(-5-1)^2+(2+1)^2}=\sqrt{(-6)^2+(3)^2}=\sqrt{36+9}=\sqrt{45}=3\sqrt5$

$\overline{AD}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(-2+5)^2+(3-2)^2}=\sqrt{(3)^2+(1)^2}=\sqrt{9+1}=\sqrt{10}$

$\overline{AE}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(-2+3)^2+(3-1)^2}=\sqrt{(1)^2+(2)^2}=\sqrt{1+4}=\sqrt{5}$

Perimeter of the parallelogram ABCD is,

$=2(\overline{AD}+\overline{CD})\\\\=2(\sqrt{10}+3\sqrt5)\\\\=19.74\ units$

Area of the parallelogram ABCD is,

$=\overline{CD}\times \overline{AE}\\\\=3\sqrt5\times \sqrt{5}\\\\=3\times 5\\\\=15\ sq.unit$

7 0

3 years ago