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almond37 [142]
3 years ago
13

A b c or d !!!! HELP FASTTT

Mathematics
1 answer:
slavikrds [6]3 years ago
3 0
C!!
hope this helped
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Help !!!!!!!!!!!!!!!!!!!!!
vladimir1956 [14]

Answer:

12 is the answer thanks for asking in

3 0
3 years ago
7 earn 5 coins
Sergio039 [100]
The answer is 9 cubic units
5 0
3 years ago
a(x+b)=4x+10 in the equation above a and b are constants. if the equation has infinitely many solutions for X, what is the value
xeze [42]

Answer:

b= \frac{5}{2}

Step-by-step explanation:

Given

a(x+b)=4x+10\\ax+b= 4x+10\\

Now we know that system has infinite solution for x

\frac{a_1}{a_2}=\frac{b_1}{b_2}

in above equation.

a_1=a, a_2=4, b_1=ab,b_2=10\\

∴\frac{a}{4}=\frac{ab}{10}\\\therefore b=\frac{10\times a}{4\times a}=\frac{5}{2}

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
Please help!!
fgiga [73]

The price of the small pots is $2.40 so you would have 2.4s ( multiply the number of small pots by the price)

She bought a total of 14 pots, so the number of large pots would be 14 - s ( subtract the number of small pots from the total )


Now you have:

L = 14-s

2.4s + 5.6(14-s) = 49.6


The answer is C.

7 0
3 years ago
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