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3 years ago

Solve the unknown variable: 5k/16 +2k/12 = (-25+k)/24

2 answers:

8 0

5k/16+2k/12=(-25+k)/24
find common denomenator so we can cancel them
find GCM of 16,12, and 24

16=2*2*2*2
12=2*2*3
24=2*2*2*3
GCM=2*2*2*2*3=48

multiply everybody by 48

3(5k)+4(2k)=2(-25+k)
distribute
15k+8k=-50+2k
add like terms
23k=-50+2k
subtract 2k
21k=-50
divie both sides by 21
k=-50/21
k=-2 and 4/25

Goryan [66]3 years ago

6 0

Hello. (60k+32k)/192=(-25+k)/24 ; 92k/192=(-25+k)/24 ; 92k×24=192×(-25+k)/24 ; 2208k= -4800+192k ; 2208k-192k= -4800 ; 2016k= -4800 ; k= -4800/2016 ; k= -2,38. I hope to have helped you.

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A rectangular room measures 18 ft by 37 ft. How many

Lera25 [3.4K]

Step-by-step explanation:

the area of a rectangle is length × width.

in our case

37 × 18 = 666 ft²

so, we need 666 ft² of tiles.

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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

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Sliva [168]

Answer:

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2 years ago

BRAINLIEST ASAP! PLEASE HELP ME :)

nikdorinn [45]

Answer:

m <1 = 61

Step-by-step explanation:

We know that the angle adjacent to 130 is 50 degrees because linear angles are complementary.

Extending line t to make a triangle, we can find the angle adjacent to the angle labeled 111 ( and above it) is 69.

That makes the angle that touches line q 61 to finish the triangle

This angle and angle 1 are corresponding angles, so they are equal

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3 years ago