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3 years ago

NEED HELP .SERIOUS ANSWERS

Mathematics

1 answer:

Maksim231197 [3]3 years ago

4 0

It is rotated 90 degress clockwise then reflected across the x-axis

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What is 2.57142857143 as an improper fraction in its simplest form?

Sidana [21]

Answer:

257142857143 / 100000000000.

Step-by-step explanation:

2.57142857143

= 2 + 57142857143 / 100,000,000,000

= 200,000,000,000 + 57142857143 / 100,000,000,000

= 257142857143 / 100000000000.

6 0

3 years ago

gregori [183]

It’s D bc Dt= 15x3 and it’s not that hard to solve

6 0

3 years ago

A bakery the price of the cake is nine dollars more than the price of a pie. One day the bakery sold 8 cakes and 14 pies for a t

gavmur [86]

Answer:

The price of the cake is $24 and the price of the Pie is $15

Step-by-step explanation:

Given

Represent price of Cake with C and Price of Pie with P

$C = 9 + P$

Cakes sold = 8

Pies sold = 14

$Total = \$402$

Required

Determine C and P

To represent the cakes and pies sold, we have the following expression

$8C + 14P = 402$

Substitute 9 + P for C

$8(9+P) + 14P = 402$

Open the bracket

$72 + 8P + 14P = 402$

Collect Like Terms

$8P + 14P = 402 - 72$

$22P= 330$

Divide both sides by 22

$P = \frac{330}{22}$

$P = 15$

Recall that

$C = 9 + P$

$C = 9 + 15$

$C = 24$

Hence, the price of the cake is $24 and the price of the Pie is $15

7 0

3 years ago

For all values of x f(x) = 3x + 2 Find f-¹(-12)

Novosadov [1.4K]

Answer:

-4.67

Step-by-step explanation:

first find f^-1(x)

let f(x)=y

y=3x+2

3x=y-2

x=y-2/3

Therefore f^-1(x)=x-2/3

Now f^-1(-12)=-12-2/3

=-14/3

=-4.67

8 0

2 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