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Arlecino [84]
3 years ago
5

tle=" \frac{3}{2} x - \frac{3}{4} x = 6" alt=" \frac{3}{2} x - \frac{3}{4} x = 6" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
MAVERICK [17]3 years ago
3 0

\frac{3}{2}x-\frac{3}{4}x=6

\frac{3}{2}x*4-\frac{3}{4}x*4=6*4

6x-3x=24

3x=24

\frac{3x}{3}=\frac{24}{3}

x=8

"x=8 is correct answer."

Hope this helps!

Thanks!

-Charlie

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What is the eighth term of the arithmetic sequence? An=<br> 10n – 14?
Alja [10]

Answer:

A8 = 66 [8th term]

Step-by-step explanation:

Given the explicit arithmetic sequence An = -14 + 10n → -4 + 10(n-1)

To find the 8th term. An nth term is where n is n in the sequence. So substitute 8 for n, and simplify.

So A8 = -4 + 10(8-1) → -4 + 10(7) → -4 + 70 → 70 - 4 → 66.

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With bases V1, V2 , V3 andw1, w2 , W3, suppose T(v1) = w2 and T(v2) = T(v3) = w1 + w3 . T is a linear transformation. Find the m
Roman55 [17]

Answer:

See picture and explanation below.

Step-by-step explanation:

With this information, the matrix A that you can find is the transformation matrix of T. The matrix A is useful because T(x)=Av for all v in the domain of T.

A is defined as T=([T(v_1)] [T(v_2] [T(v_3)])\text{ where }[T(v_i)] denotes the vector of coordinates of T(v_i) respect to the basis (we can apply this definition because forms a basis for the domain of T).

The vector of coordinates can be computed in the following way: if T(v_i)=a_1w_1+a_2w_2+a_3w_3 then [T(v_i)]=(a_1,a_2,a_3)^t.

Note that we have all the required information: T(v_1)=0w_1+1\cdot w_2+0w_3 then [T(v_1)]=(0,1,0)^t

T(v_2)=T(v_3)=1\cdot w_1+0w_2+0w_3 hence [T(v_2)]=[T(v_3)]=(1,0,0)^t

The matrix A is on the picture attached, with the multiplication A(1,1,1).

Finally, to obtain the output required at the end, use the properties of a linear transformation and the outputs given:

T(v_1+v_2+v_3)=T(v_1)+T(v_2)+T(v_3)=w_2+w_1+w_1=w_2+2w_1  

In this last case, we can either use the linearity of T or multiply by A.

4 0
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