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3 years ago

If a has to be greater than or equal to 1 and less than 10, what does a equal?

Mathematics

1 answer:

kirill115 [55]3 years ago

3 0

Answer:

i think the answer is 5

Step-by-step explanation:

if it greater than or equal to its 1 or higher but, its not 10 because it has to be less than 10 5 falls right in the middle of greater that or equal to 1 but less than 10

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For each hour he babysits, Anderson earns $1 more than half of Carey’s hourly rate. Anderson earns $6 per hour. Which equation c

Illusion [34]

Answer: Our required equation would be $6=\dfrac{c}{2}+1$

Step-by-step explanation:

Since we have given that

Amount that Anderson earns per hour = $6

According to question, Anderson earns $1 more than half of Carey's hourly rate.

Let the hourly rate of Carey be 'c'.

So, it becomes,

$6=\dfrac{c}{2}+1$

And the value of c would be

$6=\dfrac{c}{2}+1\\\\6-1=\dfrac{c}{2}\\\\5=\dfrac{c}{2}\\\\c=\$10$

Hence, our required equation would be $6=\dfrac{c}{2}+1$

5 0

2 years ago

Write an expression that is equivalent to 4n – 12 + 8y + (-3n).

Dimas [21]

It would be 1n - 12 + 8y.
4n - 3n = 1n
12 + 8y don’t mix so we can leave it.
Therefore: 1n - 12 + 8y

7 0

2 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 $\mathbb{R}^2$