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4 years ago

12

Determine whether the set, together with the indicated operations, is a vector space. if it is not, then identify at least one o

f the vector space axioms that fails.

1 answer:

lesya692 [45]4 years ago

3 0

<span>Where is The set?? can you add it please</span>

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I just need x and y please

Alenkinab [10]

Answer:

Step-by-step explanation:

Since you know both triangles are congruent, you can say that the corresponding sides are equal. For example, you can say:

17x+3 = 57

because they are corresponding sides of both triangles. To further solve the equation:

17x = 54 --> x = 54/17 --> <u>3 3/17</u>

Same thing for Y. lines PX and FD are congruent, so, you can say:

4y-3 = 54

4y=57

<u>y= 14.25</u>

6 0

3 years ago

PLEASE HELP ASAP!!! which one???

Mice21 [21]

Answer:

b

Step-by-step explanation:

8 0

3 years ago

If the simple interest on $7000 for 8 years is $4480, then what is the interest rate ?

OverLord2011 [107]

$560 for every year.

5 0

3 years ago

Assume that x and y are both differentiable functions of t and find the required values of dy/dt and dx/dt.

qwelly [4]

Answer:

A)

$\displaystyle \frac{dy}{dt}=-\frac{33}{8}$

B)

$\displaystyle \frac{dx}{dt}=\frac{3}{2}$

Step-by-step explanation:

<em>x</em> and <em>y</em> are differentiable functions of <em>t, </em>and we are given the equation:

$xy=6$

First, let's differentiate both sides of the equation with respect to <em>t</em>. So:

$\displaystyle \frac{d}{dt}\left[xy\right]=\frac{d}{dt}[6]$

By the Product Rule and rewriting:

$\displaystyle \frac{d}{dt}[x(t)]y+x\frac{d}{dt}[y(t)]=0$

Therefore:

$\displaystyle y\frac{dx}{dt}+x\frac{dy}{dt}=0$

A)

We want to find dy/dt when <em>x</em> = 4 and dx/dt = 11.

Using our original equation, find <em>y</em> when <em>x</em> = 4:

$\displaystyle (4)y=6\Rightarrow y=\frac{3}{2}$

Therefore:

$\displaystyle \frac{3}{2}\left(11\right)+(4)\frac{dy}{dt}=0$

Solve for dy/dt:

$\displaystyle \frac{dy}{dt}=-\frac{33}{8}$

B)

We want to find dx/dt when <em>x</em> = 1 and dy/dt = -9.

Again, using our original equation, find <em>y</em> when <em>x</em> = 1:

$(1)y=6\Rightarrow y=6$

Therefore:

$\displaystyle (6)\frac{dx}{dt}+(1)\left(-9)=0$

Solve for dx/dt:

$\displaystyle \frac{dx}{dt}=\frac{3}{2}$

5 0

3 years ago

Can someone please help me answer these

Anni [7]

Step-by-step explanation:

5: odd

6: even

7: even

That's your answer

3 0

3 years ago