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4 years ago

Help immediately I need answer. Plssssssss (ᗒᗣᗕ)՞

Mathematics

1 answer:

abruzzese [7]4 years ago

7 0

Answer:

490 :)

Step-by-step explanation:

Top Surface Area = 210

Bottom SA = 280

TOP + BOTTOM = 490

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Do problem 6 show work i will give brainliest if ur correct.

aliina [53]

Step-by-step explanation:

they forget the last step which is to take the square root of the sum

7) a = 4

b = 17

a² + b² = c²

4²+ 17² = c²

16 + 289 = c²

305 = c²

√305 = c

17.464249196572980 = C

7 0

3 years ago

Round 4.9451 to the nearest thousandths

maria [59]

Answer:

49.451 this would be the answer

5 0

3 years ago

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I start with 8 grapes in my yard. The grapes triple every week. In how many weeks will there be 320 grapes in my yard (rounded t

REY [17]

U think it’s 40 cuz 8 x40 equal 320 ;-; if it’s wrong my name will come to use

4 0

3 years ago

QUESTION 3 [10 MARKS] A bakery finds that the price they can sell cakes is given by the function p = 580 − 10x where x is the nu

Andru [333]

Answer:

(a)Revenue=580x-10x²,

Marginal Revenue=580-20x

(b)Fixed Cost, =900

Marginal Cost,=300+50x

(c)Profit Function=280x-900-35x²

(d)x=4

Step-by-step explanation:

The price, p = 580 − 10x where x is the number of cakes sold per day.

Total cost function,c = (30+5x)²

(a) Revenue Function

R(x)=x*p(x)=x(580 − 10x)

R(x)=580x-10x²

Marginal Revenue Function

This is the derivative of the revenue function.

If R(x)=580x-10x²,

R'(x)=580-20x

(b)Total cost function,c = (30+5x)²

c=(30+5x)(30+5x)

=900+300x+25x²

Therefore, Fixed Cost, =900

Marginal Cost Function

This is the derivative of the cost function.

If c(x)=900+300x+25x²

Marginal Cost, c'(x)=300+50x

(c)Profit Function

Profit, P(x)=R(x)-C(x)

=(580x-10x²)-(900+300x+25x²)

=580x-10x²-900-300x-25x²

P(x)=280x-900-35x²

(d)To maximize profit, we take the derivative of P(x) in (c) above and solve for its critical point.

Since P(x)=280x-900-35x²

P'(x)=280-70x

Equate the derivative to zero

280-70x=0

280=70x

x=4

The number of cakes that maximizes profit is 4.

5 0

3 years ago

An airplane climbs 100 feet during the first second after takeoff. In each succeeding second it climbs 100 feet more than it cli

tigry1 [53]

Answer:

15 seconds.

Step-by-step explanation:

∵ The distance covered by plane in first second = 100 ft,

Also, in each succeeding second it climbs 100 feet more than it climbed during the previous second,

So, distance covered in second second = 200,

In third second = 300,

In fourth second = 400,

............, so on....

Thus, the total distance covered by plane in n seconds = 100 + 200 + 300 +400......... upto n seconds

$=\frac{n}{2}(2\times 100 + (n-1)100)$ ( Sum of AP )

$=\frac{n}{2}(200+100n-100)$

$=\frac{n}{2}(100+100n)$

$=50n+50n^2$

Suppose the distance covered in n seconds is 12,000 feet,

$\implies 50n+50n^2=12000$

$n+n^2=240$

$n^2+n-240=0$

$n^2+16n-15n-240=0$

$n(n+16)-15(n+16)=0$

$(n-15)(n+16)=0$

$\implies n=15\text{ or }n=-16$

∵ n can not be negative,

Hence, after 15 seconds the plane will reach an altitude of 12,000 feet above its takeoff height.

6 0

3 years ago

Read 2 more answers