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3 years ago

Laura has $30 to spend on food for the week. Which inequality represents this? x < 30 x < 30 x > 30 x > 30

Mathematics

2 answers:

Katen [24]3 years ago

7 0

The answer is B, she cant spend more, and she can spend less than $30.

steposvetlana [31]3 years ago

6 0

In the given question Laura has $30 to spend on food for the week.

According to the question we have to set the inequality between total amount and amount spent on food.

Amount spend by Laura on food for the week = $30

Let the amount spent by Laura on food be x

So she has to spend up to $30

So inequality will be $x\leqslant 30$

So option b is the answer

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Step-by-step explanation:

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3 years ago

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krok68 [10]

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x = 9

Step-by-step explanation:

since the terms are in AP then there is a common difference between consecutive terms, that is

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2 years ago

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Degger [83]

Answer:

Step-by-step explanation:

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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

$\lambda_1 = \cdots = \lambda_n$ .

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

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

four time the first of three consecutive integers is six more than the product of two and the third integer. find the integers.

maw [93]

<span>
Assign variables :
Let x = first consecutive even
Let x + 2 = second consecutive even
Let x + 4 = third consecutive even
4x = 6 + 2(x+4)
4x = 6 + 2x + 8
4x = 2x + 14
4x - 2x = 14
so your answer is 7,9and 11</span>

3 0

3 years ago