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Oduvanchick [21]
3 years ago
14

I need help marking brainliest

Mathematics
2 answers:
Furkat [3]3 years ago
7 0
Answer: 2

Explanation: 1 makes no sense. With 3 there would still be the pollution to worry about and there’s the same problem with 4. Because there is more pollution the cricket frogs would die out.
saw5 [17]3 years ago
3 0

Answer:

2) Decreases as the habitat is increasingly polluted

Step-by-step explanation:

number 1 make no sense. 3; at the start, say the population is x. but taking into account the pesticide, it cant no longer ramain the same (pesticide kills some of the x) . 4; similar to the case for answer 3

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Without using calculator find :<br> cos(225) sin(315) + sin( –270) tan(405)
Ronch [10]

Answer:

1.5

Step-by-step explanation:

225° bisects Q III

315° bisects Q IV

-270° = 90°

405° = 45° bisects Q I

cos(225) sin(315) + sin( –270) tan(405) = ?

      (-½√2)(-½√2) + 1(1) = ?

                          0.5 + 1 = 1.5

6 0
2 years ago
Read 2 more answers
Which equation represents the function graphed on the
puteri [66]

Answer:

it is the third option of g(x)=|x-1|+3x-1

the way I look at this is by visualizing the translations using the parent function of f(x)=|x|

the function moved up three and to the right by one which led to my conclusion

6 0
3 years ago
Which pairs of numbers have the same absolute value? How are these numbers related?
NeTakaya

Answer:

4=4

Step-by-step explanation:

3 0
2 years ago
consider the quadratic form q(x,y,z)=11x^2-16xy-y^2+8xz-4yz-4z^2. Find an orthogonal change of variable that eliminates the cros
Bezzdna [24]

Answer:

q(x,y,z)=16x^{2}-5y^{2}-5z^{2}

Step-by-step explanation:

The given quadratic form is of the form

q(x,y,z)=ax^2+by^2+dxy+exz+fyz.

Where a=11,b=-1,c=-4,d=-16,e=8,f=-4.Every quadratic form of this kind can be written as

q(x,y,z)={\bf x}^{T}A{\bf x}=ax^2+by^2+cz^2+dxy+exz+fyz=\left(\begin{array}{ccc}x&y&z\end{array}\right) \left(\begin{array}{ccc}a&\frac{1}{2} d&\frac{1}{2} e\\\frac{1}{2} d&b&\frac{1}{2} f\\\frac{1}{2} e&\frac{1}{2} f&c\end{array}\right) \left(\begin{array}{c}x&y&z\end{array}\right)

Observe that A is a symmetric matrix. So A is orthogonally diagonalizable, that is to say,  D=Q^{T}AQ where Q is an orthogonal matrix and D is a diagonal matrix.

In our case we have:

A=\left(\begin{array}{ccc}11&(\frac{1}{2})(-16) &(\frac{1}{2}) (8)\\(\frac{1}{2}) (-16)&(-1)&(\frac{1}{2}) (-4)\\(\frac{1}{2}) (8)&(\frac{1}{2}) (-4)&(-4)\end{array}\right)=\left(\begin{array}{ccc}11&-8 &4\\-8&-1&-2\\4&-2&-4\end{array}\right)

The eigenvalues of A are \lambda_{1}=16,\lambda_{2}=-5,\lambda_{3}=-5.

Every symmetric matriz is orthogonally diagonalizable. Applying the process of diagonalization by an orthogonal matrix we have that:

Q=\left(\begin{array}{ccc}\frac{4}{\sqrt{21}}&-\frac{1}{\sqrt{17}}&\frac{8}{\sqrt{357}}\\\frac{-2}{\sqrt{21}}&0&\sqrt{\frac{17}{21}}\\\frac{1}{\sqrt{21}}&\frac{4}{\sqrt{17}}&\frac{2}{\sqrt{357}}\end{array}\right)

D=\left(\begin{array}{ccc}16&0&0\\0&-5&0\\0&0&-5\end{array}\right)

Now, we have to do the change of variables {\bf x}=Q{\bf y} to obtain

q({\bf x})={\bf x}^{T}A{\bf x}=(Q{\bf y})^{T}AQ{\bf y}={\bf y}^{T}Q^{T}AQ{\bf y}={\bf y}^{T}D{\bf y}=\lambda_{1}y_{1}^{2}+\lambda_{2}y_{2}^{2}+\lambda_{3}y_{3}^{2}=16y_{1}^{2}-5y_{2}^{2}-5y_{3}^2

Which can be written as:

q(x,y,z)=16x^{2}-5y^{2}-5z^{2}

4 0
3 years ago
Describe a relationship that the graph may represent
mylen [45]

Answer:

an increase then a pause then anotha increase

Step-by-step explanation:

4 0
3 years ago
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