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3 years ago

8

Colin joins a gym across the street from his house. The gym charges an initial fee of $15 for the key card plus $20 per month. W

e’rea relationship between total cost,C and number of months,m.

1 answer:

arsen [322]3 years ago

5 0

C=20m+15 represents the relationship between cost and number of months.

Step-by-step explanation:

Given,

Initial fee charged by gym = $15

Per month fee of gym = $20

Let,

m be the number of months.

C be the total cost for m months.

Total cost = Initial fee + Per month fee*number of months

C = 15+20m

C=20m+15

C=20m+15 represents the relationship between cost and number of months.

Keywords: equation, addition

Learn more about addition at:

#LearnwithBrainly

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The volume of this regular pentagonal pyramid is 82.5 cubic meters. What is the height of the pyramid?

a_sh-v [17]

the height of the pentagonal pyramid is 5. 70 meters

<h3>Volume of a regular pentagonal pyramid</h3>

The formula for determining the volume of a regular pentagonal pyramid is given as;

V=5/12tan(54°)ha^2

Where

a is the base edge
h is the height

We have the volume to be;

volume = 82. 5 cubic centers

height = h

a = 5m

Substitute the values

$82. 5 = \frac{5}{12}$ × $tan 54$ × $h$ × $5^2$

$82. 5 = 0. 42$ × $1. 3764$ × $25$ × $h$

Make 'h' subject of formula

$h = \frac{82. 5}{14. 45}$

h = 5. 70 meters

The height of the pentagonal pyramid is 5. 70 meters

Thus, the height of the pentagonal pyramid is 5. 70 meters

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6 0

1 year ago

Paper rectangle ABCD is folded in such a way that the fold passes through point B, and so A coincides with point H where H ∈ CD.

aivan3 [116]

The answer is BC = 38.22 cm.

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

We have, ∠BKD = 120° ,BK = 28 cm, Draw a perpendicular from point K on BC let it intersect at point M. In right angled ΔBMK, ∠BKM=30° and BK = 28 cm

sin30° = perpendicular/hypotenuse

1/2 = BM/BK

1/2 = BM/28

BM= 14 cm

Now , In right angled ΔBMK ,

cos30° = base/hypotenuse

√3/2 = MK/28

MK = 14√3 = 24.22 cm

KMCD is a square MK = MC = 24.22 cm

also, BC = BM + MC , putting values of BM & MC we get :

BC = 14 cm + 24.22 cm

BC = 38.22 cm.

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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

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zhannawk [14.2K]

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A box contains 421 books. If your teacher orders 19 boxes of books, about how many books will arrive? Use estimation to find you

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