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crimeas [40]
3 years ago
6

Please Help! (10 POINTS)

Mathematics
2 answers:
Sliva [168]3 years ago
5 0
The initial measure is:
 m1 = 28 ft
 The real measure must be:
 m2 = 14 ft
 The correct procedure to determine the scale factor is:
 k = m1 / m2

k = 28/14
 Dividing both numbers between 14 we have:
 k = (28/14) / (14/14)
 Rewriting:
 k = 2/1

k = 2
 Answer:
 
Zelie should have divided both numbers by 14.
creativ13 [48]3 years ago
3 0

Answer:

<h2>it is A.</h2>

Step-by-step explanation:

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What is the measure of Zc<br>​
maks197457 [2]

Answer: 56 degrees

Step-by-step explanation:

We know that angle between b and c is 90 degrees. Because there is a line dividing angles a and b from 124 and c, we also know that each side is 180 degrees (a and b add to 180, and 124 and c add to 180).

124 and c are supplementary angles. We can represent this in an equation to solve for c:

124 + c = 180\\\\c = 56

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Need help with this. Please
seropon [69]

Answer: hey im not sure, but i think your supposed to solve 7x-6 and 9+12x. Add these two together and times it by 55. Im not syre if this is right but u could try it.

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A cake recipe calls for 4.5 deciliters of milk. How many liters of milk is needed for the recipe?
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23. Starting with x=0.81818181 , express x as a fraction in simplest form.
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3 years ago
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Assume {v1, . . . , vn} is a basis of a vector space V , and T : V ------&gt; 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

8 0
3 years ago
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