Ask question

Ask question

All categories

3 years ago

7

PLEASE HELP ASAP!!! I NEED CORRECT ANSWERS ONLY PLEASE!!! I NEED TO FINISH THESE QUESTIONS BEFORE MIDNIGHT TONIGHT.

1 answer:

Lina20 [59]3 years ago

7 0

Yo sup??

we can solve this question by applying trigonometric ratios

cos59=CB/CD

CD=CB/cos59

=7.8

Hope this helps.

You might be interested in

If C(x) is the cost of producing x units of a commodity, then the average cost per unit is c(x) = C(x)/x. Consider the cost func

kvasek [131]

Answer:

Step-by-step explanation:

Given that:

If C(x) = the cost of producing x units of a commodity

Then;

then the average cost per unit is c(x) = $\dfrac{C(x)}{x}$

We are to consider a given function:

$C(x) = 54,000 + 130x + 4x^{3/2}$

And the objectives are to determine the following:

a) the total cost at a production level of 1000 units.

So;

If C(1000) = the cost of producing 1000 units of a commodity

$C(1000) = 54,000 + 130(1000) + 4(1000)^{3/2}$

$C(1000) = 54,000 + 130000 + 4( \sqrt[2]{1000^3} )$

$C(1000) = 54,000 + 130000 + 4(31622.7766)$

$C(1000) = 54,000 + 130000 + 126491.1064$

$C(1000) = $310491.1064$

$\mathbf{C(1000) \approx $310491.11 }$

(b) Find the average cost at a production level of 1000 units.

Recall that :

the average cost per unit is c(x) = $\dfrac{C(x)}{x}$

SO;

$c(x) =\dfrac{(54,000 + 130x + 4x^{3/2})}{x}$

Using the law of indices

$c(x) =\dfrac{54000}{x} + 130 + 4x^{1/2}$

$c(1000) = \dfrac{54000}{1000}+ 130 + {4(1000)^{1/2}}$

c(1000) =$ 310.49 per unit

(c) Find the marginal cost at a production level of 1000 units.

The marginal cost is C'(x)

Differentiating C(x) = 54,000 + 130x + 4x^{3/2} to get C'(x) ; we Have:

$C'(x) = 0 + 130 + 4 \times \dfrac{3}{2} \ x^{\dfrac{3}{2}-1}$

$C'(x) = 0 + 130 + 2 \times \ {3} \ x^{\frac{1}{2}}$

$C'(x) = 0 + 130 + \ {6}\ x^{\frac{1}{2}}$

$C'(1000) = 0 + 130 + \ {6} \ (1000)^{\frac{1}{2}}$

$C'(1000) = 319.7366596$

$\mathbf{C'(1000) = \$319.74 \ per \ unit}$

(d) Find the production level that will minimize the average cost.

the average cost per unit is c(x) = $\dfrac{C(x)}{x}$

$c(x) =\dfrac{54000}{x} + 130 + 4x^{1/2}$

the production level that will minimize the average cost is c'(x)

differentiating $c(x) =\dfrac{54000}{x} + 130 + 4x^{1/2}$ to get c'(x); we have

$c'(x)= \dfrac{54000}{x^2} + 0+ \dfrac{4}{2 \sqrt{x} }$

$c'(x)= \dfrac{54000}{x^2} + 0+ \dfrac{2}{ \sqrt{x} }$

Also

$c''(x)= \dfrac{108000}{x^3} -x^{-3/2}$

$c'(x)= \dfrac{54000}{x^2} + \dfrac{4}{2 \sqrt{x} } = 0$

$x^2 = 27000\sqrt{x}$

$\sqrt{x} (x^{3/2} - 27000) =0$

x= 0; or $x= (27000)^{2/3}$ = $\sqrt[3]{27000^2}$ = 30² = 900

Since production cost can never be zero; then the production cost = 900 units

(e) What is the minimum average cost?

the minimum average cost of c(900) is

$c(900) =\dfrac{54000}{900} + 130 + 4(900)^{1/2}$

c(900) = 60 + 130 + 4(30)

c(900) = 60 +130 + 120

c(900) = $310 per unit

8 0

3 years ago

The dryers at Quik-Wash Laundry cost a quarter for 7 minutes. How many quarters will you have to use to dry your clothes for 45

Alchen [17]

It would not be an exact amount of quarters, but you would need 7 quarters in total to dry your clothes for 45 minutes. All you have to do is divide 45 by 7, which is roughly 6.42 quarters.

4 0

3 years ago

1/1.5 = 1/2.0 + 1/c<br><br><br><br> find c and please please show all your work

Arturiano [62]

1/1.5=1/2+1/c
1/1.5–1/2=1/c
(2–1.5)/3=1/c
0.5/3=1/c
0.5c/0.5=3/0.5
C=6

6 is the answer. I hope this helps.

4 0

3 years ago

HELP URGENT ASAP WILL MARK AND 5 STARS DUE in 5 MINUETS HURRY

mel-nik [20]

Answer:

1. x=8

the other ones i got was decimals and i dont think you want those for this assignment

Step-by-step explanation:

7 0

3 years ago

If 5x+3y=12 and 4x-5y=17, what is the value of y?

shepuryov [24]

Answer:

i d k

Step-by-step explanation:

6 0

3 years ago