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AveGali [126]
3 years ago
14

Write a function defined by y = ƒ(x), subject to the condition that the value of ƒ(x) is 16 times the square root of x.

Mathematics
1 answer:
Bess [88]3 years ago
7 0

16 times the square root of x:

    16  *           √x                                  so f(x) = 16√x

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The producer of certain medicine claims that it’s bottling
aliya0001 [1]

Answer:

it depends if he's  bottling it or not

Step-by-step explanation:

4 0
3 years ago
Solve the equation y + 5=-12 <br> A y=-19<br> B y=-17<br> C y=-7<br> D y=7
mrs_skeptik [129]

Answer:

y=17

Hope this helps you

8 0
3 years ago
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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
Simplify the expression.
Nana76 [90]
The answer is the A. I hope this helps!
6 0
3 years ago
3(x+1)-5=3x-2<br><br> Option1: no solution<br> Option2:infinite solutions<br> Option3:one solution
Vikentia [17]

Answer:

Simplify 3(x+1)−5.

3x−2=3x−2

Move all terms containing x to the left side of the equation.

−2=−2 Since −2=−2 , the equation will always be true for any value of x

All real numbers

The result can be shown in multiple forms.

All real numbers

Interval Notation:

(−∞,∞)

Option2 : infinite solutions

Step-by-step explanation:

3 0
3 years ago
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