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3 years ago

Find two numbers, if … c Their sum is 3/5 and their difference is 7 1/5

Mathematics

1 answer:

Makovka662 [10]3 years ago

7 0

Answer:

39/10 and - 33/10.

Step-by-step explanation:

7 1/5 = 36/5 so if the 2 numbers are x and y, we have:

x + y = 3/5

x - y = 36/5

Adding the 2 equations to eliminate y:

2x = 39/5

x = 39/10.

So 39/10 + y = 3/5 (from the first equation).

y = 3/5 - 39/10

y = 6/10 - 39/10

y = -33/10.

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1- 4/5 divided by 2/3

Softa [21]

I hope this helps you

7 0

3 years ago

The coordinates of polygon ABCD are A(-4,-1), B(-2,3), C(2,2), and D(4,-3). Use these coordinates to complete the sentences belo

oee [108]

Answer:

The perimeter of polygon ABCD, to the nearest thousandth units is

22.227 units

The perimeter of polygon ABCD', to the nearest thousandth, would be 20.980 units

The area of polygon ABCD' is 19.5 units²

Step-by-step explanation:

The coordinates forming the polygon are

A (-4,-1), B(-2,3), C(2,2), and D(4,-3)

The perimeter then is given by the sum of the length of the sides as follows;

Length of line between two X and Y points distance, xi and yj apart is

length XY = $\sqrt{x^2+y^2}$

Therefore, the length between points AB is

length AB = $\sqrt{(-4 - (-2))^2+(-1-3)^2}$

= $\sqrt{(-4 +2)^2+(-4)^2} = \sqrt{-2^2+-4^2} = \sqrt{20}$ = 4.472 units

Similarly, length BC is given by

Length BC = $\sqrt{(-4)^2+1^2} = \sqrt{17}$ = 4.123 units

Length CD = $\sqrt{(-2)^2+5^2} = \sqrt{29}$ = 5.385 units

Length DA = $\sqrt{(8)^2+(-2)^2} = \sqrt{68}$ = 8.246 units

The perimeter is equal to;

length AB + Length BC + Length CD + Length DA

= 4.472 + 4.123 + 5.385 + 8.246 = 22.227 units

If the point D is moved up 2 units and left 1 unit we have

D' = x-1, y+2 where x and y are the coordinates of point D

D(4, -3) → D'((4-1), (-3+2)) = D'(3, -1)

The length D'A = $\sqrt{(7)^2+(0)^2} = \sqrt{49}$ =7 units

The perimeter of polygon ABCD'

length AB + Length BC + Length CD + Length D'A

= 4.472 + 4.123 + 5.385 + 7 = 20.980 units.

The area is given by the determinant of the 3 by 3 matrix using Cramer's Rule as follows

$S_{\bigtriangleup}$ = $(\frac{1}{2})|x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2|$

= $(\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|$

Triangle ABC we have

A(-4,-1)

B(-2,3)

C(2,2)

$S_{{\bigtriangleup}ABC}$ = $(\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|$

$=(\frac{1}{2})|(-4)(3-2)+(-2)(2-(-1)) +2((-1)-3)|$

= $=(\frac{1}{2})|-4-6 -8|= 9 units^2$

Triangle ACD'

A(-4,-1)

C(2,2)

D'(3, -1)

$S_{{\bigtriangleup}ACD'}$ = $(\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|$

$=(\frac{1}{2})|(-4)(2+1)+(2)(-1+1) +3((-1)-2)| = \frac{21}{2}$ = 10.5 units²

Area of polygon = $S_{{\bigtriangleup}ABCD'}$ = $S_{{\bigtriangleup}ABC}$ + $S_{{\bigtriangleup}ACD'}$ = (9 + 10.5) units²

Area of polygon ABCD' = 19.5 units².

7 0

3 years ago

SUppose the total cost C(x) to manufacture a quantity x of insecticide (in hundreds of liters) is given by

Hitman42 [59]

Answer:

a) $C(x)$ is increasing in two regions: (i) $(+\infty, 8\,s)$ and (ii) $(10\,s,+\infty)$ .

b) $C(x)$ decreases in $(8\,s, 10\,s)$ .

Step-by-step explanation:

Let $C(x) = x^{3}-27\cdot x^{2}+240\cdot x +850$ , where $x$ is the quantity of insecticide, measured in hundreds of liters, and $C(x)$ is the total manufacturing cost as a function of the quantity of the insecticide, measured in US dollars. A possible approach to determine which regions of $C(x)$ are decreasing and increasing by means of the first derivative and graphing tools. The first derivative of the function is:

$C'(x) = 3\cdot x^{2}-54\cdot x+240$ (1)

Please notice that regions where C(x) is increasing has $C'(x) > 0$ , whereas $C'(x) < 0$ when $C(x) < 0$ .

We notice that $C(x)$ is increasing in two regions: (i) $(+\infty, 8\,s)$ and (ii) $(10\,s,+\infty)$ . Besides, $C(x)$ decreases in $(8\,s, 10\,s)$ .