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rosijanka [135]
3 years ago
15

NOTE: ONLY FOR EXPERTS.... ONLY

Mathematics
1 answer:
jekas [21]3 years ago
5 0
Wow, you are right. This is for experts only lol
You might be interested in
Nadia says the hypotenuse of the right triangle has a length of 73 because the Pythagorean theorem states that (28+45)squared eq
Setler79 [48]

Answer: Her solution is incorrect. The hypotenuse is 53 units

Step-by-step explanation: The error occurred when she quoted the Pythagorean theorem. She added up both sides and squared them and made the answer equal to the square of the hypotenuse. She was supposed to have squared them first, before adding up.

The Pythagorean (or Pythagoras) theorem states that the sum of the squares of two sides of a right angled triangle equals the square of the hypotenuse of the triangle. That is, in triangle ABC, if the hypotenuse is given as line AC, the Pythagoras theorem states that

AC^2 = AB^2 + BC^2

Where AC is the hypotenuse and AB and BC are the other two sides.

In Nadia’s solution, if the hypotenuse is given as 73, and the other two sides are 28 and 45, then the Pythagorean theorem properly expressed would be as follows;

73^2 = 28^2 + 45^2

5329 = 784 + 2025

5329 ≠ 2809

As we have determined, writing out the expression as

73^2 = (28 + 45)^2 is very incorrect as this is not the Pythagorean theorem and of course, the triangle cannot have 73 as its hypotenuse as we have determined by properly writing out the Pythagorean theorem.

The correct solution is as follows;

AC^2 = AB^2 + BC^2

AC^2 = 28^2 + 45^2

AC^2 = 784 + 2025

AC^2 = 2809

Add the square root sign to both sides of the equation

AC = 53

Therefore, the hypotenuse is 53 units

4 0
3 years ago
Read 2 more answers
What is the length of the diagonal of a 10 cm by 15 cm rectangle?
Ksivusya [100]

Answer:

18.02 cm

Step-by-step explanation:

A diagonal makes a rectangle into two right triangles. So if we use the Pythagorean theorem we can find the hypotenuse which is the diagonals.

Remember, because we know the length and width, we know the two sides.

10^2 + 15^2 = c^2

100+225 = c^2

325 = c^2

18.027 ≈ c

4 0
3 years ago
Read 2 more answers
I don’t understand :(
viktelen [127]

Answer:

The quotient when x^3-4x^2-8x+8 is divided by x+2 is x^2-6x+4 with no remainder, so x+2 is a factor of p(x).

Step-by-step explanation:

As the question suggests, there are two ways to solve this problem: long division (which can always be used to divide polynomials), and synthetic division (which can only be used when you are dividing by something of the form (x-a). In this case, we may use synthetic division, since x+2 is equal to x-(-2). I will use synthetic division here as it is slightly faster than long division.

The -2 on the left represents that we are dividing by x-(-2), and the rest of the numbers in the top row are the coefficients of p(x): 1, -4, -8, and 8.

The first step in synthetic division is to being down the leading coefficient of the polynomial: in this case, the 1 (as indicated in red). Now, we multiply the 1 by -2. We get -2, which we place directly below the -4.

Next, we add directly down the column, (-4 + (-2) is equal to -6), and this answer is placed in the box below. We can continue this process, getting the coefficients 1, -6, 4, and 0 in the bottom row.

This is the answer: the quotient is x^2-6x+4, and the remainder is zero (which indicates that x+2 is a factor of p(x)).

--

Edit: long division

We may also solve this problem using long division (see the second image). The first step is to look at the leading coefficients: since x \times x^2 is equal to x^3, the first term in the quotient will be x^2. Since x^2 \times (x+2) is equal to x^3+2x^2, we must now subtract that from x^3-4x^2-8x+8.

We repeat the same process, as shown in the image. Since we eventually get to zero, the remainder is zero, and the polynomial at the top (x^2-6x+4) is the quotient.

<em>I know this might be difficult to follow, so please comment if you have any questions.</em>

4 0
3 years ago
Suppose the following number of defects has been found in successive samples of size 100: 6, 7, 3, 9, 6, 9, 4, 14, 3, 5, 6, 9, 6
Brut [27]

Answer:

Given the data in the question;

Samples of size 100: 6, 7, 3, 9, 6, 9, 4, 14, 3, 5, 6, 9, 6, 10, 9, 2, 8, 4, 8, 10, 10, 8, 7, 7, 7, 6, 14, 18, 13, 6.

a)

For a p chart ( control chart for fraction nonconforming), the center line and upper and lower control limits are;

UCL = p" + 3√[ (p"(1-P")) / n ]

CL = p"

LCL = p" - 3√[ (p"(1-P")) / n ]

here, p" is the average fraction defective

Now, with the 30 samples of size 100

p" =  [∑(6, 7, 3, 9, 6, 9, 4, 14, 3, 5, 6, 9, 6, 10, 9, 2, 8, 4, 8, 10, 10, 8, 7, 7, 7, 6, 14, 18, 13, 6.)] / [ 30 × 100 ]

p" = 234 / 3000

p" = 0.078

so the trial control limits for the fraction-defective control chart are;

UCL = p" + 3√[ (p"(1-P")) / n ]

UCL = 0.078 + 3√[ (0.078(1-0.078)) / 100 ]

UCL = 0.078 + ( 3 × 0.026817 )

UCL = 0.078 + 0.080451

UCL = 0.1585

LCL = p" - 3√[ (p"(1-P")) / n ]

LCL = 0.078 - 3√[ (0.078(1-0.078)) / 100 ]

LCL = 0.078 - ( 3 × 0.026817 )

LCL = 0.078 - 0.080451

LCL =  0 ( SET TO ZERO )

Diagram of the Chart uploaded below

b)

from the p chart for a) below, sample 28 violated the first western electric rule,

summary report from Minitab;

TEST 1. One point more than 3.00 standard deviations from the center line.

Test failed at points: 28

Hence, we conclude that the process is out of statistical control

Lets Assume that assignable causes can be found to eliminate out of control points.

Since 28 is out of control, we should eliminate this sample and recalculate the trial control limits for the P chart.

so

p" = 0.0745

UCL = p" + 3√[ (p"(1-P")) / n ]

UCL = 0.0745 + 3√[ (0.0745(1-0.0745)) / 100 ]

UCL = 0.0745 + ( 3 × 0.026258 )

UCL = 0.0745 + 0.078774

UCL = 0.1532

LCL  = p" - 3√[ (p"(1-P")) / n ]

LCL = 0.0745 - 3√[ (0.0745(1-0.0745)) / 100 ]

LCL = 0.0745 - ( 3 × 0.026258 )

LCL = 0.0745 - 0.078774

UCL = 0  ( SET TO ZERO )

The second p chart diagram is upload below;

NOTE; the red circle symbol on 28 denotes that the point is not used in computing the control limits

7 0
2 years ago
If you are using Cosine and need to solve for an unknown hypotenuse, you should...
kifflom [539]

Using cosine, we can find the hypotenuse by using the formula below;

hypotenuse = adjacent / cos ∅

<h3 /><h3 /><h3>Trigonometric ratios:</h3>
  • Trigonometric ratios are ratios of sides of a right angled triangle.
  • The simplest ratios are cosine, sine and tangent.

Using cosine, the hypotenuse side can be solved as follows:

cosine ∅ = adjacent / hypotenuse

cross multiply

hypotenuse cos ∅ = adjacent

divide both sides by cos ∅

hypotenuse cos ∅ / cos ∅ = adjacent / cos ∅

Therefore,

hypotenuse = adjacent / cos ∅

learn more on cosines here: brainly.com/question/10657732?referrer=searchResults

6 0
2 years ago
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