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3 years ago

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The perimeter of a square is 4 times as great as the length of any of its sides. Determine if the perimeter of a square is propo

rtional to its side length

2 answers:

stiv31 [10]3 years ago

7 0

P= 4 times L so yes. Hope it helps!

klemol [59]3 years ago

6 0

<span> Perimeter is proportional to the side length by Perimeter, which 4 times the sides which equals 1/4 Perimeter.</span>

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Corbin made a scale model of the San Jacinto Monument. The monument has an actual height of 604 feet. Corbin's model used a scal

photoshop1234 [79]

<u>Answer: </u><u>6.04 inches</u>

<u>Step-by-step explanation:</u>

The actual height of the monument = 604 feet

The scale used by Corbins is 1 inch = 100 feet

The height of the Corbins model in inches= $\frac{604}{100}$ = 6.04 inches

the height of the Corbins model = 6.04 inches

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1 year ago

What is the slope of ( 5,1 ) and ( 2,-7 )

PolarNik [594]

Answer:

8/3 is the slope

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3 years ago

How to solve mean median and mode

cestrela7 [59]

You can solve the median by putting all the numbers in order. Then cross off one at a time at the start, then cross off the one at the end, then keep going until you get to one.

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8 0

3 years ago

△ABC≅△DEF, BC=4x−12, and EF=−3x+16. Find x and EF.

kirill [66]

x = 4 and EF = 4 ⇒ 3rd answer

Step-by-step explanation:

If two triangles are congruent, then

1. Their corresponding sides are equal

2. Their corresponding angles are equal

3. Their areas and perimeters are equal

∵ △ ABC ≅ △ DEF

∴ AB ≅ DE

∴ BC ≅ EF

∴ AC ≅ DF

∵ BC = 4 x - 12

∵ EF = -3 x + 16

∵ BC = EF

∴ 4 x - 12 = -3 x + 16

Let us solve the equation to find x

∵ 4 x - 12 = -3 x + 16

- Add 3 x for both sides

∴ 7 x - 12 = 16

- Add 12 to both sides

∴ 7 x = 28

- Divide both sides by 7

∴ x = 4

Substitute the value of x in the expression of EF

∵ EF = -3 x + 16

∵ x = 4

∴ EF = -3(4) + 16 = -12 + 16

∴ EF = 4

x = 4 and EF = 4

Learn more:

You can learn more about congruence of Δs in brainly.com/question/3202836

#LearnwithBrainly

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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 $\mathbb{R}^2$

8 0

3 years ago