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Fittoniya [83]
3 years ago
12

Need help please help me

Mathematics
1 answer:
Anna35 [415]3 years ago
4 0

Answer:

4

Step-by-step explanation:

This is a piece wise function. A piece wise function has 2 or more functions which only work on specific intervals. To find when x = -2, first find the interval that contains x = -2.

The interval is x < 0. This interval corresponds to the function x².

So f(-2) will be (-2)² = 4

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igor_vitrenko [27]
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4 0
2 years ago
Solve by factorization<br> 2x^2 + 3x -20 = 0
saul85 [17]

Answer:

x=2.5 or x=-4

Step-by-step explanation:

factorisation= factors:1, 2, 4, 5, 10, 20 (only the positive factors)

We will use the factors 4 and 5.

=(2x-5)(x+4)=0

2x-5=0   || x+4=0

2x=+5      || x=-4

x=2.5  or x=-4

4 0
2 years ago
Read 2 more answers
The data in which table represents a linear function that has a slope of zero? A 2-column table with 5 rows. Column 1 is labeled
lisov135 [29]

Answer:   A 2-column table with 5 rows. Column 1 is labeled x with entries negative 5, negative 4, negative 3, negative 2, negative 1. Column 2 is labeled y with entries 5, 5, 5, 5, 5.

8 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
A line passes through the point (4,9) and has a slope of 5/2.
kkurt [141]

Answer:

5/2

Step-by-step explanation:

BRAINLIEST

8 0
2 years ago
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