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Anarel [89]
3 years ago
7

Ted is a single guy who’s living the good life. The spreadsheet below shows Ted’s cash flow for a month.

Mathematics
1 answer:
svet-max [94.6K]3 years ago
3 0
Its not there, the picture.
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Find KL. Round to the nearest hundredth.<br> d<br> Sut of<br> .1.3<br> 48<br> L
VARVARA [1.3K]

Answer:

1.3

Step-by-step explanation:

give brainliest

4 0
3 years ago
I will mark brainliest please helppp
Ilia_Sergeevich [38]

Step-by-step explanation:

The x intercept is the point that touches the x axis

The y-intercept is the point that touches the y-axis

So...

The x-intercept = (0, -40)

The y-intercept = (0, 15)

5 0
3 years ago
A lawn mower is pushed a distance of 100 ft. Along a horizontal path by a constant force of 60 lbs. Yhe handle of the lawn mower
stira [4]
For this case we have the following equation:
 w = F • PQ
 Where,
 w: work done
 F: is the force vector
 PQ: is the vector of the direction of movement.
 Rewriting the equation we have:
 w = || F || • || PQ || costheta
 Substituting values:
 w = (60) * (100) * (cos (45))
 w = (60) * (100) * (root (2) / 2)
 w = 4242.640687 lb.ft
 Answer:
 
The work done pushing the lawn mower is:
 
w = 4242.6 lb.ft
7 0
3 years ago
You made two deposits to your bank account this month. One deposit was $24.61, and the second deposit was $16.49. Your balance a
Rina8888 [55]

The <em><u>correct answer</u></em> is:

$72.31 – $24.61 – $16.49; $31.21

Explanation:

We know we end the month with $72.31.

We made two deposits during the month and no other activity.  This means if we take the amounts of the deposits away from the total at the end of the month, we can find how much we had at the beginning of the month:

72.31-24.61-16.49 = 31.21

6 0
3 years ago
Read 2 more answers
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
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