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Anuta_ua [19.1K]
2 years ago
8

PLEASE HELP ITS MATH!!!!!

Mathematics
1 answer:
LenKa [72]2 years ago
7 0

Answer:

The first one, yes they are similar, they have at least two corresponding angles

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The hypotenuse of an isosceles right triangle is centimeters longer than either of its legs. Find the exact length of each side.
Varvara68 [4.7K]

Question

The hypotenuse of an isosceles right triangle is 11 centimeters longer than either of its legs. find the exact length of each side.

Answer:

Hypotenuse:37.56

Other\ Legs: 26.56

Step-by-step explanation:

Let the hypotenuse be y and the other legs be x.

So:

y = 11 + x

Required

Determine the exact dimension of the triangle

Using Pythagoras theorem;

y^2 = x^2 + x^2

y^2 = 2x^2

This gives:

(11+x)^2 = 2x^2

Open bracket

121 + 22x + x^2= 2x^2

Collect like terms

2x^2 - x^2 -22x - 121 = 0

x^2 -22x - 121 = 0

Using quadratic formula:

x = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}

x = \frac{-(-22)\±\sqrt{(-22)^2 - 4*1*-121}}{2*1}

x = \frac{-(-22)\±\sqrt{968}}{2*1}

x = \frac{22\±31.11}{2}

Split

x = \frac{22+31.11}{2}\ or\ x = \frac{22-31.11}{2}\\

x = \frac{53.11}{2}\ or\ x = \frac{-9.11}{2}

x can not be negative.

So:

x = \frac{53.11}{2}

x = 26.56

Recall that:

y = 11 + x

y = 11 + 26.56

y = 37.56

Hence, the dimensions are:

Hypotenuse:37.56

Other\ Legs: 26.56

6 0
3 years ago
1. Iris has two ribbons. The first ribbon is 11.7 in. long. The second ribbon is 5.25 in. long. She wants to cut them so that th
lisov135 [29]
The estimated lengths are 12in. and 5 in. This would be an estimated difference of 7in.


The estimated difference is going to be higher than the actual difference because in rounding 11.7 to 12, you are losing .30in, but in rounding down 5.25 to 5, you are actually gaining .75in.


The ACTUAL difference is 6.45in
3 0
3 years ago
What number can be written as 40 plus 5 ?
Sergio [31]
40+5=45 is written in extended form so your answer is 45
5 0
3 years ago
Read 2 more answers
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:

<u>Part A </u>

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

<u>Part B </u>

<u>(Fixed Cost)</u>

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 .

<u> Marginal Cost Function</u>

If  c=900+300x+25x^2

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

<u>Part C </u>

<u>Profit Function </u>

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

<u> Part D </u>

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
3. Problem 2: Suppose a radio manufacturer has the total cost function C (x) = 43x + $ 1850 and the revenue function R (x) = $ 8
OLEGan [10]

Given that,

Total cost function, C (x) = 43x + $1850

The revenue function R (x) = $80x

To find,

The number of units that must be produced and sold to break even.

Solution,

At break even, cost = revenue

43x + $1850 = $80x

Subtract 80x from both sides.

43x + 1850 -80x = $80x -80x

1850 = 80x-43x

37x = 1850

x = 50

So, the required number of units are 50.

8 0
3 years ago
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