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3 years ago

1. Classify the equation 7x + 3 = 7x – 4 as having one solution, no solution, or

Mathematics

2 answers:

vampirchik [111]3 years ago

7 0

Answer:

no solution.

Step-by-step explanation:

1. 7x + 3 = 7x - 4 Subtract 7x on both sides

2. 3 = -4

Three does not equal to negative four so, it has no solutions

Nana76 [90]3 years ago

3 0

The answer is no solution.

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Help in the question above please

Shkiper50 [21]

Answer:

A)11

Step-by-step explanation:

These are matrices one dimensional with one column and 3 rows each.

-The product of the matrices is obtained by multiplying the correspond values and summing up;

$pq=\left[\begin{array}{ccc}3\\2\\-1\end{array}\right] \times\left[\begin{array}{ccc}5\\-1\\2\end{array}\right] \\\\\\\\=(3\times 5)+(2\times -1)+(-1\times 2)\\\\=15+-2+-2\\\\=11$

Hence, the product of p and q is 11

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3 years ago

Jorge's hourly salary is $7.65. last week he worked 23 hour week how much did he earn

lions [1.4K]

$7.65 × 23 = $175.95

Answer: $175.95

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3 years ago

Simplify the expressions

4vir4ik [10]

Answer:

1) 2(2x+3)

2) 2x+4

3) - X + 4

4) 9x - 2

5) 5x−2

6) - 12x - 17

7) - 15x + 11

8) 3x/4

9) - 4x + 7

10) 4x-26/2

11) - 8x + 20.8

12) x - 6

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2 years ago

What is the approximate length of the radius of a circle with a circumference of 63 inches?

Sergeeva-Olga [200]

Answer: 63/2π

Step-by-step explanation: C=πd

63=πd

D=63/π

r=1/2d

1/2×63/π=<u>63/2π</u>

5 0

3 years ago

Assume {v1, . . . , vn} is a basis of a vector space V , and T : V ------> W is an isomorphism where W is another vector spac

Degger [83]

Answer:

Step-by-step explanation:

To prove that $w_1,\dots w_n$ form a basis for W, we must check that this set is a set of linearly independent vector and it generates the whole space W. We are given that T is an isomorphism. That is, T is injective and surjective. A linear transformation is injective if and only if it maps the zero of the domain vector space to the codomain's zero and that is the only vector that is mapped to 0. Also, a linear transformation is surjective if for every vector w in W there exists v in V such that T(v) =w

Recall that the set $w_1,\dots w_n$ is linearly independent if and only if the equation

$\lambda_1w_1+\dots \lambda_n w_n=0$ implies that

$\lambda_1 = \cdots = \lambda_n$ .

Recall that $w_i = T(v_i)$ for i=1,...,n. Consider $T^{-1}$ to be the inverse transformation of T. Consider the equation

$\lambda_1w_1+\dots \lambda_n w_n=0$

If we apply $T^{-1}$ to this equation, then, we get

$T^{-1}(\lambda_1w_1+\dots \lambda_n w_n) =T^{-1}(0) = 0$

Since T is linear, its inverse is also linear, hence

$T^{-1}(\lambda_1w_1+\dots \lambda_n w_n) = \lambda_1T^{-1}(w_1)+\dots + \lambda_nT^{-1}(w_n)=0$

which is equivalent to the equation

$\lambda_1v_1+\dots + \lambda_nv_n =0$

Since $v_1,\dots,v_n$ are linearly independt, this implies that $\lambda_1=\dots \lambda_n =0$ , so the set $\{w_1, \dots, w_n\}$ is linearly independent.

Now, we will prove that this set generates W. To do so, let w be a vector in W. We must prove that there exist $a_1, \dots a_n$ such that

$w = a_1w_1+\dots+a_nw_n$

Since T is surjective, there exists a vector v in V such that T(v) = w. Since $v_1,\dots, v_n$ is a basis of v, there exist $a_1,\dots a_n$ , such that

$a_1v_1+\dots a_nv_n=v$

Then, applying T on both sides, we have that

$T(a_1v_1+\dots a_nv_n)=a_1T(v_1)+\dots a_n T(v_n) = a_1w_1+\dots a_n w_n= T(v) =w$

which proves that $w_1,\dots w_n$ generate the whole space W. Hence, the set $\{w_1, \dots, w_n\}$ is a basis of W.

Consider the linear transformation $T:\mathbb{R}^2\to \mathbb{R}^2$ , given by T(x,y) = T(x,0). This transformations fails to be injective, since T(1,2) = T(1,3) = (1,0). Consider the base of $\mathbb{R}^2$ given by (1,0), (0,1). We have that T(1,0) = (1,0), T(0,1) = (0,0). This set is not linearly independent, and hence cannot be a base of $\mathbb{R}^2$

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3 years ago