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dedylja [7]
2 years ago
15

Whats the answer to this numeric algebraic expression

Mathematics
1 answer:
Degger [83]2 years ago
4 0

Answer:

a+b+2

Step-by-step explanation:

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A. Sketch the curve by using the parametric equations to plot points. Indicate with an arrow the direction in which the curve is
Furkat [3]

Answer:

y  =  2 -  ( x - 1)²/9

Step-by-step explanation: See Annex

As the parameter t increases,  the value of x increases and the value of y decreases, we get the figure of the annex ( the arrow in the Annex indicates the way of the curve with t increasing.

b) to eliminate the parameter:

x  =  1  + 3*t    (1)          y  =  2  -  t²  (2)

Then  from equation (1)  t  = ( x  - 1 ) / 3

plugging that value in equation (2)

y  =  2  -  [ (  x  -  1 ) / 3 ]²

y  =  2  - ( x² + 1 - 2*x)/9

9*y  =  18  - x² + 1 - 2*x    or    y  =  2 -  ( x - 1)²/9

The curve is a parabol

Download pptx
5 0
3 years ago
Seventy-five students were in the audience
aleksley [76]
75/200 movie goers watching 3D were students. 37.5% were students
50/100 movie goers watching 2D were students. 50% were students.
7 0
2 years ago
Prove that if {x1x2.......xk}isany
Radda [10]

Answer:

See the proof below.

Step-by-step explanation:

What we need to proof is this: "Assuming X a vector space over a scalar field C. Let X= {x1,x2,....,xn} a set of vectors in X, where n\geq 2. If the set X is linearly dependent if and only if at least one of the vectors in X can be written as a linear combination of the other vectors"

Proof

Since we have a if and only if w need to proof the statement on the two possible ways.

If X is linearly dependent, then a vector is a linear combination

We suppose the set X= (x_1, x_2,....,x_n) is linearly dependent, so then by definition we have scalars c_1,c_2,....,c_n in C such that:

c_1 x_1 +c_2 x_2 +.....+c_n x_n =0

And not all the scalars c_1,c_2,....,c_n are equal to 0.

Since at least one constant is non zero we can assume for example that c_1 \neq 0, and we have this:

c_1 v_1 = -c_2 v_2 -c_3 v_3 -.... -c_n v_n

We can divide by c1 since we assume that c_1 \neq 0 and we have this:

v_1= -\frac{c_2}{c_1} v_2 -\frac{c_3}{c_1} v_3 - .....- \frac{c_n}{c_1} v_n

And as we can see the vector v_1 can be written a a linear combination of the remaining vectors v_2,v_3,...,v_n. We select v1 but we can select any vector and we get the same result.

If a vector is a linear combination, then X is linearly dependent

We assume on this case that X is a linear combination of the remaining vectors, as on the last part we can assume that we select v_1 and we have this:

v_1 = c_2 v_2 + c_3 v_3 +...+c_n v_n

For scalars defined c_2,c_3,...,c_n in C. So then we have this:

v_1 -c_2 v_2 -c_3 v_3 - ....-c_n v_n =0

So then we can conclude that the set X is linearly dependent.

And that complet the proof for this case.

5 0
3 years ago
Which statements are true?
lyudmila [28]

Answer:

the true statement is

All parallelograms are quadrilaterals

Step-by-step explanation:

5 0
3 years ago
Please help me quick ASAP!!
Crank
A hope this helps 
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7 0
3 years ago
Read 2 more answers
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