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Art [367]
3 years ago
13

A is between C and D. True or false?

Mathematics
1 answer:
DanielleElmas [232]3 years ago
3 0

Answer:

false

Step-by-step explanation:

false , because A is between C and B , it must be on the same straight line

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What is the ordinate of the point where the graph of the<br> equation y = 5x + 2 crosses the y-axis?
eimsori [14]

Answer:

(0,2)

Step-by-step explanation:

in any y = mx + b formula, b is always the y intercept, and the y intercept is just a fancy name for where the line first crosses the y axis. hope this helps!

8 0
2 years ago
What is the equation of the axis of symmetry?
Aleksandr [31]

Answer:

Step-by-step explanation:

8 0
2 years ago
Paige’s income statement for the month of December is shown. Paige monthly income statement for December. Total wages are 1,850
DedPeter [7]

Answer:

Paige’s net income for December is $950

Step-by-step explanation:

Here in this question, we are interested in calculating Paige’s net income for December

Mathematically;

Net income = Monthly wage - Total Expenses

From the question, monthly wage = $1,850 while the total expenses = $900

Thus, the net income will be ; $1,850 - $900 = $950

6 0
3 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
Please help out thanksssss yallllllll!!!!!!!!!
Mila [183]

Answer:

okyyyyyyyyyyyyyyyyyy

4 0
2 years ago
Read 2 more answers
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