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3 years ago

Help with 2 plz I really need help

Mathematics

1 answer:

Kazeer [188]3 years ago

4 0

the whole triangle is call an acute equilibrium triangle if that's the answer your looking for if not I need to see the whole question

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Maitri and Aabhas do a work in 12 hours.

tamaranim1 [39]

Answer:

See below ~

Step-by-step explanation:

Given

Maitri and Aabhas do a work in 12 hours
Aabhas and Kavya do the work in 15 hours
Kavya and Maitri do the work in 20 hours

Solving

Take Maitri, Aabhas, and Kavya to be x, y, z respectively
x + y = 12 (1)
y + z = 15 (2)
x + z = 20 (3)

Take Equation 1 and rewrite it so that it is equal to x.

x = 12 - y

Take Equation 2 and rewrite it so that it is equal to z.

z = 15 - y

Now, substitute these values in Equation 3.

x + z = 20
12 - y + 15 - y = 20
-2y + 27 = 20
2y = 7
y = 7/2 = 3.5 hours [Aabhas]

Substitute the value of y in Equation 1.

x + 3.5 = 12
x = 8.5 hours [Maitri]

Substitute the value of y in Equation 2.

3.5 + z = 15
z = 11.5 hours [Kavya]

Add the values of x, y, and z together.

x + y + z
8.5 + 3.5 + 11.5
12 + 11.5
23.5 hours [together]

8 0

2 years ago

(−2)⋅(8)−(4)⋅(−3) A. -17 B. -3 C. -4

Gwar [14]

The Answer To This Problem is: C.-4

8 0

3 years ago

A rectangular box is to have a square base and a volume of 12 ft3. If the material for the base costs $0.17/ft2, the material fo

katen-ka-za [31]

Answer:

(a)Length =2 feet

(b)Width =2 feet

(c)Height=3 feet

Step-by-step explanation:

Let the dimensions of the box be x, y and z

The rectangular box has a square base.

Therefore, Volume of the box $V=x^2z$

Volume of the box $=12 ft^3\\$

$Therefore, x^2z=12\\z=\frac{12}{x^2}$

The material for the base costs $\$0.17/ft^2$ , the material for the sides costs $\$0.10/ft^2$ , and the material for the top costs $\$0.13/ft^2$ .

Area of the base $=x^2$

Cost of the Base $=\$0.17x^2$

Area of the sides $=4xz$

Cost of the sides= $=\$0.10(4xz)$

Area of the Top $=x^2$

Cost of the Base $=\$0.13x^2$

Total Cost, $C(x,z) =0.17x^2+0.13x^2+0.10(4xz)$

Substituting $z=\frac{12}{x^2}$

$C(x) =0.17x^2+0.13x^2+0.10(4x)(\frac{12}{x^2})\\C(x)=0.3x^2+\frac{4.8}{x} \\C(x)=\dfrac{0.3x^3+4.8}{x}$

To minimize C(x), we solve for the derivative and obtain its critical point

$C'(x)=\dfrac{0.6x^3-4.8}{x^2}\\Setting \:C'(x)=0\\0.6x^3-4.8=0\\0.6x^3=4.8\\x^3=4.8\div 0.6\\x^3=8\\x=\sqrt[3]{8}=2$

Recall: $z=\frac{12}{x^2}=\frac{12}{2^2}=3\\$

Therefore, the dimensions that minimizes the cost of the box are:

(a)Length =2 feet

(b)Width =2 feet

(c)Height=3 feet

7 0

3 years ago

Ms. chen is getting a new tile floor for her workshop. Her workshop is 15ft long and 11ft wide.

ANTONII [103]

A cause you just gotta think about it and realize you got it

3 0

3 years ago

Jason went shopping.

melomori [17]

53.55 / 85 = 0.63

0.63 * 100 = 63

63 - 38 = 25

watch was 25 before discount

8 0

2 years ago