A120 watt light bulb uses about 0.1 kilowatt of electricity per hour. if electricity costs $0.20 per kilowatt hour how much does it cost to have a bulb on for an hour?

$0.02 is the cost to have the bulb on for an hour

Step-by-step explanation:

A 120 watt light bulb uses about 0.1 kilowatt of electricity per hour

We can assume 1 kilowatt per hour = 1 unit

So total units of electricity consumed in 1 hour = 0.1 × 1

= 0.1 kilowatt hour or 0.1 unit

Since cost of one unit or 1 kilowatt hour is = $0.20 per unit or kilowatt hour.

Therefore, total cost to have this bulb on for an hour = Number of units consumed by the light bulb × per unit cost

= 0.1×0.20

= $0.02

$0.02 is the cost to have the bulb on for an hour.

x² + 11x + 18 < 0  

[-11 ± sqrt(121 - 72)]/2 = (-11 ± 7)/2  

solution: (-9, -2)

step-by-step explanation:

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)