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KengaRu [80]
4 years ago
6

Bonnie buys a used car for $2500. She calculates her average monthly expenses for gas, insurance, and maintenance at $150 per mo

nth. The function f(x) = 150x + 2500 represents her total cost of owning the car for x number of months.
Mathematics
1 answer:
zhannawk [14.2K]4 years ago
6 0
F(x) = 150x + 2500

The range is f(x) ≥ 2500
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Help me please, I’ll give brainliest!!
polet [3.4K]

Answer:

A line that is perpendicular to y=2/5x-5 must have a slope of -5/2, but a line parallel to line y=5/2x+6 must have a slope of 5/2, and these slopes do not match up.

Explanation:

In order for a line to be parallel to another line, they must have the same slope. In order for a line to be perpendicular to another line, their slopes must be opposite reciprocals (meaning that when they are multiplied, the product is -1). From here you can find what slope the line must have to perpendicular or parallel to another line, and then you can see that the answers to not match up to be the same line!

Hope this helped :)

5 0
3 years ago
The average value of y=v(x) equals 4 for 1≤x≤6, and equals 5 for 6≤x≤8. what is the average value of v(x) for 1≤x≤8 ?
Tomtit [17]
The average value over the interval is the area under the curve divided by the width of the interval.

Area = 4*(6 -1) +5(8 -6) = 30
Width = 8 - 1 = 7

Average value = 30/7 = 4 2/7

7 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
Help please!!!!!!!!!!!!!!!! Asap
Hunter-Best [27]
One positive and one negative
6 0
3 years ago
The purchase price of an iPad is $625.00. There is a 15% discount and the tax rate is
vagabundo [1.1K]

Answer:

$573.75

Step-by-step explanation:

625.00x15%=93.75

625.00-93.75=531.25

531.25x8%=42.50

531.25+42.50=573.75

5 0
3 years ago
Read 2 more answers
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