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Triss [41]
3 years ago
10

Please help me with number 1 a

Mathematics
1 answer:
Ronch [10]3 years ago
4 0
The answer i got was they cost the same
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14 over 15 rounded to the nearest tenth
NARA [144]

Answer:

0.9

Step-by-step explanation:

14/15 = 0.933333333  <u>IF:</u>

<u></u>

<u>1.</u><em> You first change it into a decimal by dividing the numerator by the denominator.</em>

<em>        </em>14 ÷ 15

<u>2.</u> <em>You will get a recurring decimal.</em>

<em>      </em>0.933333333

<u>3.</u><em> The first digit after the decimal is in the position of the '</em><u><em>tenths'</em></u>

      0.<u>9</u>

<u>4.</u><em> You should use only two digits after the decimal to calculate </em>

<em>       0.93</em>

<u>5.</u><em> Since the number 3 is less than five, it will change into a zero and get eliminated.</em>

<em>             ∴ The answer will be = 0.9</em>

5 0
3 years ago
Workout Gym has a sign up fee of $15 and charges $1 per day for membership.
Papessa [141]

Answer:

The second graph is the correct graph

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Please do this ASAP! Due tomorrow. 9:12pm here.
Mumz [18]

Answer:

compare (1/2)^3 is greater than (a/b)^2

Step-by-step explanation:

1/2*1/2*1/2= 0.125 or 1/8

1/4 * 1/4=0.0625 or 1/16

8 0
3 years ago
Geomtry plzzzz help me
victus00 [196]

Step-by-step explanation:

it is an isoceles triangle (equally long legs).

there is a right angle at W, so YW is the height of the isoceles triangle XYZ.

that means W splits XZ in half.

therefore,

ZW = XZ/2 = 38/2 = 19

8 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
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