the production of a printer consists of the cost of raw material at 100 dollars the cost of overheads at 80$ and wages at 120$ if the cost of raw materials and overheads are increased by 11% and 20% respectively while wages are decreased by 15% find the percentage increase or decrease in the production cost of the printer

Answers

Answer 1

Answer:

Answer:

The percentage increase in the production cost of the printer is 3%.

Step-by-step explanation:

We are given that the production of a printer consists of the cost of raw material at 100 dollars the cost of overheads at 80$ and wages at 120$.

Also, the cost of raw materials and overheads are increased by 11% and 20% respectively while wages are decreased by 15%.

Cost of raw material = $100

Cost of overheads = $80

Cost of wages = $120

So, the total cost of the printer = $100 + $80 + $120

= $300

Now, the increase in the cost of raw material = $100 + 11% of $100

= $\$100 + ((11)/(100) * \$100)$

= $100 + $11 = $111

The increase in the cost of overheads = $80 + 20% of $80

= $\$80 + ((20)/(100) * \$80)$

= $80 + $16 = $96

The decrease in the cost of wages = $120 - 15% of $120

= $\$120 - ((15)/(100) * \$120)$

= $120 - $18 = $102

So, the new cost of a printer = $111 + $96 + $102 = $309

Now, the percentage increase in the production cost of the printer is given by;

% increase = $\frac{\text{Net increase in the cost of printer}}{\text{Original cost of printer}} * 100$

= $(\$309- \$300)/(\$300) * 100$

= 3%

Hence, the percentage increase in the production cost of the printer is 3%.

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Write the equation of the line in slope-intercept form that passes through (3, 10) and (2, 4)

Answers

Answer:

y=6x-8

Step-by-step explanation:

10-4/3-2

m=6/1

y=mx+b

y=6x+b

4=6(2)+b

4=12+b

-8=b

y=6x-8

What is a solution to 3/4 a is greater than -16

Answers

Answer:

a > - 21.33

a > -21 1/3

Step-by-step explanation:

3/4a > -16

a > -16*4/3

a > - 64/3

a > - 21.33

a > -21 1/3

A trough has a semicircular cross section with a radius of 9 feet. Water starts flowing into the trough in such a way that the depth of the water is increasing at a rate of 2 inches per hour. (a) Give a function w = f(t) relating the width w of the surface of the water to the time t, in hours. Make sure to specify the domain and compute the range too.(b) After how many hours will the surface of the water have width of 6 feet?

(c) Give a function t = f −^1 (w) relating the time to the width of the surface of the water. Make sure to specify the domain and compute the range too.

Answers

Answer:

(a) Let h represents the height of water and w represents the width of the water,

Since, the depth of the water is increasing at a rate of 2 inches per hour,

So, after t hours,

The height of water, h(t) = 2t inches = t/6 ft,

( ∵ 1 foot = 12 inches ⇒ 1 inch = 1/12 ft )

Thus, the distance distance from the centre to the top of the water, d = 9 - h(t) ( see in the diagram )

$d=9-(t)/(6)$ ,

By the Pythagoras theorem,

$d^2 + ((w)/(2))^2 = 9^2$

$(9-(t)/(6))^2 +(w^2)/(4) = 81$

$(t^2)/(36)-(18t)/(6) + (w^2)/(4)=0$

$(t^2 - 108t + 9w^2)/(36)=0$

$t^2 - 108t + 9w^2 =0$

$9w^2 = 108t - t^2$

$w = (1)/(3)√(108t - t^2)$

Since, diameter of the semicircular cross section is 18 ft,

So, 0 ≤ w ≤ 18,

i.e Range = [0, 18]

Also, w will be defined if 108t - t² ≥ 0

⇒ (108 - t)t ≥ 0,

⇒ 0 ≤ t ≤ 108

i.e Domain = [0, 108]

(b) If w = 6,

$6 =(1)/(3)√(108t - t^2)$

$18 =√(108t-t^2)$

$324 = 108t - t^2$

$\implies t^2 - 108t+ 324=0$

By using quadratic formula,

$\implies t = 3.088\text{ or }t = 104.912$

Hence, After 3.1 hours or 104.9 hours will the surface of the water have width of 6 feet.

(c) $w = (1)/(3)√(108t- t^2)$

$\implies 3w = √(108t- t^2)$

$9w^2 = 108t - t^2$

$-9w^2 = -108t + t^2$

$-9w^2 + 2916 = 2916 - 108t + t^2$

$2916 - 9w^2 = (t - 108)^2$

$(t-108) = √(2916 - 9w^2)$

$t = √(2916 - 9w^2) + 108$

For 0 ≤ w ≤ 18,

0 ≤ t ≤ 108,

So, Domain = [0, 18]

Range = [0, 108]

Final answer:

The width of the water's surface in a semicircular trough can be represented by the function w=t/3 and its domain is t ≥ 0 and the range is 0 ≤ w ≤ 6. To have a 6 feet wide surface, thus, it would take 18 hours. The inverse function is t=3w, with a domain of 0 ≤ w ≤ 6 and range of t ≥ 0.

Explanation:

Given that the depth of the water is increasing at a rate of 2 inches per hour in a semicircular trough, we can convert this rate to feet per hour by dividing by 12, getting an increase of 1/6 feet per hour.

(a) We can express the width of the surface of the water as a function of time. We consider the cross-section of the trough is a semicircle. So, the radius of the water's surface will be the height of water, and this height increases at 1/6 feet per hour. Therefore, the width of the surface of water, w=2r=2*1/6t=t/3. The domain of the function is t ≥ 0 and the range is 0 ≤ w ≤ 6.

(b) We set w=6 in the function w=t/3 and solve for t. We get t=3*6=18 hours.

(c) The inverse function of w=t/3 is t=3w. The domain of the inverse function is 0 ≤ w ≤ 6 and the range is t ≥ 0.

Learn more about Mathematical functions here:

brainly.com/question/30594198

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What is a variable term in mathematics?

Answers

A variable is a symbol for a unknown number. It does not have a fixed number. In fact, it could be any number until you find out what it is in a equation. Or, you can never find out what it is. Yes, it is conserved a term. Hope this helped!

If you make $30 every 10 seconds how much would you make in an hour?