Im really clueless please help! no links please :(

Mathematics

2 answers:

allochka39001 [22]3 years ago

6 0

The length of BC=3 units the length of AD= 7 units the height is 3 units the area is 15 units

Zina [86]3 years ago

3 0

Height is 3 units area is 15 units

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Find the value of x

Jlenok [28]

Answer:

Hello! answer: x = 4

Step-by-step explanation:

Because this is a 180 degree angle it will add up to 180 so 8 × 4 = 32 and 32 + 120 = 152 and 7 × 4 = 28 and 28 + 152 = 180 so 4 is x HOPE THAT HELPS!!

7 0

3 years ago

Read 2 more answers

A division of Ditton Industries manufactures the Futura model microwave oven. The daily cost (in dollars) of producing these mic

Fofino [41]

Answer:

The actual cost incurred in manufacturing:

the 111st oven is $114.07
the 221st oven is 122.71
the 311st oven is $140.59

The marginal cost when x = 110, 220, and 310 is

$C'(110)=\$114.06$

$C'(220)=\$122.64$

$C'(310)=\$140.46$

Step-by-step explanation:

We know that the daily cost function of producing microwave ovens is given by

$C(x)=0.0002x^3 - 0.06x^2 + 120x + 5000$

(a) The actual cost incurred in manufacturing the 111st oven is the difference between the total cost incurred in manufacturing the first 111 ovens and the total cost of manufacturing the first 110 ovens:

$C(111)-C(110)=[0.0002(111)^3 - 0.06(111)^2 + 120(111) + 5000]-[0.0002(110)^3 - 0.06(110)^2 + 120(110) + 5000]\\\\C(111)-C(110)=17854.27-17740.2\\\\C(111)-C(110)=114.07$

For the 221st

$C(221)-C(220)=[0.0002(221)^3 - 0.06(221)^2 + 120(221) + 5000]-[0.0002(220)^3 - 0.06(220)^2 + 120(220) + 5000]\\\\C(221)-C(220)=30748.31-30625.6\\\\C(221)-C(220)=122.71$

For the 311st

$C(311)-C(310)=[0.0002(311)^3 - 0.06(311)^2 + 120(311) + 5000]-[0.0002(310)^3 - 0.06(310)^2 + 120(310) + 5000]\\\\C(311)-C(310)=42532.79-42392.2\\\\C(311)-C(310)=140.59$

(b) If C(x) is a total cost function, then the marginal cost function is defined to be its derivative C'(x).

We need to find the derivative of the total cost function:

$\frac{d}{dx}C(x)=\frac{d}{dx}(0.0002x^3 - 0.06x^2 + 120x + 5000)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\\frac{d}{dx}C(x)=\frac{d}{dx}\left(0.0002x^3\right)-\frac{d}{dx}\left(0.06x^2\right)+\frac{d}{dx}\left(120x\right)+\frac{d}{dx}\left(5000\right)\\\\C'(x)=0.0006x^2-0.12x+120$