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disa [49]
2 years ago
13

A hiker is descending 152m every 8 minutes what will be hikers change in elevation in half an hour? what as an integers​

Mathematics
1 answer:
Anon25 [30]2 years ago
4 0

Answer:

570 meters

Step-by-step explanation:

\frac{8}{152} :\frac{30}{y}

y × 8 = 152 × 30

8y = 4560

8y ÷ 8 = 4560 ÷ 8

y = 570

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Find the exact value of cos210 using the unit circle
Eduardwww [97]
Cos 210° = cos ( 180° + 30° ) - cos 30° = - √3/2
3 0
3 years ago
Verify that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial
Mariulka [41]

Answer:

i) Since P(2), P(-1) and P(½) gives 0, then it's true that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial.

ii) - the sum of the zeros and the corresponding coefficients are the same

-the Sum of the products of roots where 2 are taken at the same time is same as the corresponding coefficient.

-the product of the zeros of the polynomial is same as the corresponding coefficient

Step-by-step explanation:

We are given the cubic polynomial;

p(x) = 2x³ - 3x² - 3x + 2

For us to verify that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial, we will plug them into the equation and they must give a value of zero.

Thus;

P(2) = 2(2)³ - 3(2)² - 3(2) + 2 = 16 - 12 - 6 + 2 = 0

P(-1) = 2(-1)³ - 3(-1)² - 3(-1) + 2 = -2 - 3 + 3 + 2 = 0

P(½) = 2(½)³ - 3(½)² - 3(½) + 2 = ¼ - ¾ - 3/2 + 2 = -½ + ½ = 0

Since, P(2), P(-1) and P(½) gives 0,then it's true that 2,-1 and 1⁄2 are the zeroes of the cubic polynomial.

Now, let's verify the relationship between the zeros and the coefficients.

Let the zeros be as follows;

α = 2

β = -1

γ = ½

The coefficients are;

a = 2

b = -3

c = -3

d = 2

So, the relationships are;

α + β + γ = -b/a

αβ + βγ + γα = c/a

αβγ = -d/a

Thus,

First relationship α + β + γ = -b/a gives;

2 - 1 + ½ = -(-3/2)

1½ = 3/2

3/2 = 3/2

LHS = RHS; So, the sum of the zeros and the coefficients are the same

For the second relationship, αβ + βγ + γα = c/a it gives;

2(-1) + (-1)(½) + (½)(2) = -3/2

-2 - 1½ + 1 = -3/2

-1½ - 1½ = -3/2

-3/2 = - 3/2

LHS = RHS, so the Sum of the products of roots where 2 are taken at the same time is same as the coefficient

For the third relationship, αβγ = -d/a gives;

2 * -1 * ½ = -2/2

-1 = - 1

LHS = RHS, so the product of the zeros(roots) is same as the corresponding coefficient

7 0
3 years ago
Which rational expression is equal to a positive whole number when evaluated?
trasher [3.6K]
8^1/3 8 is 2^3 so 8^1/3 = 2 The other expresssions don't give whole numbers.
4 0
3 years ago
Read 2 more answers
a stock broker finds a bank stock that turns his $10000 investment into approximately $12000 in 2 years. over the two years what
PolarNik [594]

the your answer is 1000

4 0
3 years ago
Mathematics, 17.03.2021 23:50 burners
atroni [7]

9514 1404 393

Answer:

  1. 10 study all three
  2. 8 study only math and chem

Step-by-step explanation:

1. Adding the numbers of those who study a given subject counts ...

  • once -- those who study only one subject
  • twice -- those who study exactly two subjects
  • three times -- those who study all three subject.

That total is 40 +48 +44 = 132. When we subtract from this the total number of students, we get 132 -80 = 52, the number who study exactly 2 subjects plus twice the number who study all three.

From this, we can subtract the number who study exactly 2 subjects, leaving twice the number who study all three. 52 -32 = 20. So, 10 students study all three subjects.

__

2. Adding the numbers of those who study physics and another subject, we have 20+24 = 44. Subtracting twice the number who study all three gives the total of those how study only two subjects, one of which is physics. That total is 44 -20 = 24. So, of those 32 who study exactly two subjects, the remaining 32 -24 = 8 must be students who study only math and chemistry.

_____

It may be helpful to see the matrix of equations the problem statement describes. If we use the variables m, p, c, mp, mc, pc, mpc in that order to represent numbers of 1-subject, 2-subject, and 3-subject students, the augmented matrix looks like this.

  \left[\begin{array}{ccccccc|c}1&1&1&1&1&1&1&80\\0&1&0&1&0&1&1&40\\1&0&0&1&1&0&1&48\\0&0&1&0&1&1&1&44\\0&0&0&1&0&0&1&20\\0&0&0&0&0&1&1&24\\0&0&0&1&1&1&0&32\end{array}\right]

The solution is m=20, p=6, c=12, mp=10, mc=8, pc=14, mpc=10. The bold values are the ones this question is asking about.

3 0
3 years ago
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