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3 years ago

Please answer seriously thanks !! will give brainliest

Mathematics

1 answer:

deff fn [24]3 years ago

4 0

Answer:

Step-by-step explanation:

21) CD - chord (segment)

22)FG - line

23) EC - segment - diameter

24) AB - segment - radius

25) H - Point

26) A - Point

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After all markdowns and discounts, Charlene's prom dress cost her $22 before tax. The dress was on a rack labeled "50% off lowes

uysha [10]

Answer: $88.

Step-by-step explanation:

given data:

The cost of the dress for the prom = $22

label of the rack the dresss was on = 50%.

lowest marked price for the dress = 75%.

Solution:

if the lowest marked price of the cloth was already 75%.

and the current cost of the clothe is $22

the original cost of the cloth would be

100% – 75% = 0.25%

= $22 * 100 / 0.25%

= $88.

The original cost of the dress was $88

7 0

4 years ago

Solve the system of equations:

iVinArrow [24]

$(-1,-2) \: \: \: \: \: and \: \: \: \: \: (3,6) \\ Answer: A$

Step-by-step explanation:

$y = 2x \\ y = {x}^{2} - 3 \\ \\ 2x = {x}^{2} - 3 \\ {x}^{2} - 2x - 3 = 0 \\ (x + 1)(x - 3) = 0 \\ x + 1 = 0 \: \: \: \: \: or \: \: \: \: \: x - 3 = 0 \\ x = - 1 \: \: \: \: \: or \: \: \: \: \: x = 3$

When x = -1

Substitute x = -1 into y = 2x

y = 2(-1)

y = -2

( -1, -2)

When x = 3

Substitute x = 3 into y = 2x

y = 2(3)

y = 6

( 3, 6)

5 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

Harman [31]

Answer:

(a)Revenue function, $R(x)=580x-x^2$

Marginal Revenue function, R'(x)=580-2x

(b)Fixed cost =900 .

Marginal Cost Function=300+50x

(c)Profit, $P(x)=-35x^2+280x-900$

(d)x=4

Step-by-step explanation:

Part A

Price Function $= 580 - 10x$

The revenue function

$R(x)=x\cdot (580-10x)\\R(x)=580x-x^2$

The marginal revenue function

$\dfrac{dR}{dx}= \dfrac{d}{dx}(R(x))=\dfrac{d}{dx}(580x-x^2)=580-2x\\R'(x)=580-2x$

Part B

(Fixed Cost)

The total cost function of the company is given by $c=(30+5x)^2$

We expand the expression

$(30+5x)^2=(30+5x)(30+5x)=900+300x+25x^2$

Therefore, the fixed cost is 900 .

Marginal Cost Function

If $c=900+300x+25x^2$

Marginal Cost Function, $\frac{dc}{dx}= (900+300x+25x^2)'=300+50x$

Part C

Profit Function

Profit=Revenue -Total cost

$580x-10x^2-(900+300x+25x^2)\\580x-10x^2-900-300x-25x^2\\$Profit,P(x)=-35x^2+280x-900$

Part D

To maximize profit, we find the derivative of the profit function, equate it to zero and solve for x.

$P(x)=-35x^2+280x-900\\P'(x)=-70x+280\\-70x+280=0\\-70x=-280\\$Divide both sides by -70\\x=4$

The number of cakes that maximizes profit is 4.

6 0

3 years ago

Which of the following points are solutions to the system of inequalities shown below?

vlada-n [284]

The following solutions to the given system of inequalities shown above would be option c : (-3, -5) option d : (0, 4) option e. (4, 4) and option f. (2, -1)
So how did I know the answer. To check this, you just have to substitute the values.
For example let us take option c (-3, -5)
y < 3x + 5
-5 < 3(-3) + 5
-5 < -9 + 5
-5 < -4 <<You see that -4 is greater than -5 which makes this inequality correct. This is the same process as with the other correct options.
Hope that this answer helps.

7 0

3 years ago

Evaluate y = 2x + 1 when x= -1

Stolb23 [73]

Y = 2x + 1
y = 2(-1) + 1
y= -2 + 1
y= -1

5 0

3 years ago

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