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guapka [62]
3 years ago
15

After walking 1/4 mile from home and Han is 1/3 of his way to school. What is the distance between his home in school right mult

iplication and division equation represents the situation?
Mathematics
1 answer:
PilotLPTM [1.2K]3 years ago
5 0

Answer:

Distance from home to school =\frac{1}{4}\times 3= \frac{3}{4}\ mile

Step-by-step explanation:

Let the distance from home to school be = x miles

After walking \frac{1}{4} mile, Han is \frac{1}{3}rd of the way.

\frac{1}{3} \textrm{ of the total distance } = \frac{1}{4}\ mile

\frac{1}{3}x= \frac{1}{4}\ mile

Multiplying both sides by 3

3\times\frac{1}{3}x= \frac{1}{4}\times 3  

x= \frac{3}{4}\ mile

Distance from home to school =\frac{3}{4}\ mile

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kolbaska11 [484]

Answer:

The answer is -22

Step-by-step explanation:

remove the parentheses then you have 11 * -2 = -22

I am happy to assist :)

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4 years ago
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HURRY QUICKLY PLEASE William measures the length of three sides of a building. Place the lengths in order from greatest to least
insens350 [35]

Answer:

104m, 0.096km, 2100cm

Step-by-step explanation:

0.096 kilometres = 9600

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104m = 10400 cm

10400, 9600, 2100

104m, 0.096km, 2100cm

Answered by Gauthmath

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3 years ago
Meranda works for 7 hours at 10.25 per hour how much does she earn ?
frosja888 [35]
You take how much she makes and multiply it by the hours she works
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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

8 0
3 years ago
An electronics store has two options for liquidating televisions that have not sold.
Oksanka [162]
Answser: first option <span>350(0.95)^x + 30x - 350

Explanation:

You can figure out the </span><span>function that shows the difference in price between option 1 and option 2 if you make a table simulating the behavior for a few months:

month                 price as per option 1          price as per option 2

start                     350                                     350

1                          350 - 5% (350) =                350 - 30
                            = 350 (0,95)

2                          350 (0,95)×(0,95) =
                            350 (0,95)²                         350 - 30(2)

3                           350 (0,95)³                        350 - 30(3)

So now you can figure out the price with each option after x months:

                             350 (0.95)ˣ                        350 - 30x

And the difference is 350 (0.95)ˣ - [350 - 30x]

Which, expanding the square brackets, is 350 (0.95)ˣ + 30x - 350 ↔ the first option.
</span>
7 0
3 years ago
Read 2 more answers
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