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UNO [17]
3 years ago
7

12x-y when x= 1/3, y=-2

Mathematics
2 answers:
Darina [25.2K]3 years ago
7 0
Your equation is 13(1/3)-(-2)=x. You just solve normally. Answer: 6.3333 or 6&1/3
Basile [38]3 years ago
4 0

Answer:

Step-by-step explanation:

12 x 1/3-2

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1. Determine the measure of the unknown angles indicated by letters. Justify your answers with
Maurinko [17]

Answer:

hello,

Step-by-step explanation:

a)

In an isocele triangle, base's angles have the measure:

42+2a=180

2a=180-42

a=69(°)

b)

in a triangle, an external angle has for measure the sum of the angles not adjacents.

55+b=132

b=77 (°)

c)

in a quadrilater the sum of the (interior) angles is 2*180=360 degrees.

90+90+68+c=360

c=360-90-90-68

c=112 (°)

7 0
3 years ago
Find the length of AB.
ira [324]

Answer:

We know that in a triangle, if two angles have equal measures, then the sides opposite to them are equal in length. Thus, the length of side AB is 6 units

8 0
2 years ago
Please Help practice #3
algol [13]

Answer:

(2x+5)(4x+3)

Step-by-step explanation:

Given the expression 8x^2 + 26x + 15

Factorize

8x^2 + 26x + 15

= 8x^2 + 6x + 20x + 15

= 2x(4x+3) + 5 (4x+3)

= (2x+5)(4x+3)

Hence the factored form of the expression is (2x+5)(4x+3)

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%285%20-%20x%29%28x%20-%203%29" id="TexFormula1" title="(5 - x)(x - 3)" alt="(5 - x)(x - 3)" a
professor190 [17]

Answer:

15x

Step-by-step explanation:

because i multiply i side by side and copy the variable

3 0
3 years ago
Read 2 more answers
A grain silo has the shape of a right circular cylinder surmounted by a hemisphere. If the silo is to have a volume of 505π ft3,
Romashka-Z-Leto [24]

Answer:

radius   x  = 3 ft

height   h = 23,8  ft

Step-by-step explanation:

From problem statement

V(t)  = V(cylinder) + V(hemisphere)

let x be radius of base of cylinder (at the same time radius of the hemisphere)

and h the height of the cylinder, then:

V(c)  = π*x²*h       area of cylinder = area of base + lateral area

                                              A(c)  = π*x²  +   2*π*x*h

V(h) = (2/3)*π*x³   area of hemisphere   A(h)  =   (2/3)*π*x²

A(t)  =  π*x²  +   2*π*x*h +   (2/3)*π*x²

Now A as fuction of x    

total volume   505  = π*x²*h  +  (2/3)*π*x³

h = [505 - (2/3)* π*x³ ]  /2* π*x    

Now we have the expression for A as function of x

A(x) =  3π*x² + 2π*x*h     A(x) = 3π*x² + 505  - (2/3)π*x³

Taking derivatives both sides

A´(x) =  6πx -  2πx²              A´(x) =  0         6x  - 2x²  = 0

x₁  =  0  we dismiss

6 - 2x = 0

x = 3     and   h = [505 - (2/3)* π*x³]/2* π*x

h  =  (505 - 18.84) / 6.28*3

h  = 23,8  ft

7 0
3 years ago
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