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2 years ago

At Trendy Tailor Boutique's annual end-of-season sale, every necktie in the

Mathematics

1 answer:

astra-53 [7]2 years ago

5 0

By direct calculation, we will see that the full price of each necktie is $29.

<h3>How to find the complete price of each necktie?</h3>

Let's say that the full price of each necktie is f.

We know that Will each necktie costed $8 less than the full price, so he paid:

(f - $8) for each one.

Knowing that he bought 7 and paid a total of $147, we know that he paid:

$147/7 = $21

Then we need to solve:

(f - $8) = $21

Solving for f, we have:

f = $21 + $8 = $29.

So the full price of each necktie is $29.

If you want to learn more about algebra, you can read:

brainly.com/question/4344214

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In the orthonormal system (O; i⃗,j⃗) below, consider point A(3, −1), vector v⃗(2,1), and

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Step-by-step explanation:

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2 years ago

23. Assuming the green is flat, what is the radius of the green?

tensa zangetsu [6.8K]

Answer:

Step-by-step explanation:

$r^{2} +32^{2} = (r+8)^{2}$

$r^{2} +1024=r^{2} +16r + 64$

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2 years ago

Which represents the solution(s) of the graphed system of equations, y = x2 – 2x and y = –2x – 1? (1, –1) (0, 0) and (0, –1) (0,

olganol [36]

ANSWER

No solution

EXPLANATION

The first equation is

$y = {x}^{2} - 2x$

and the second equation is

$y = - 2x - 1$

We equate the two equations to obtain;

${x}^{2} - 2x = - 2x - 1$

This implies that

${x}^{2} = - 2x + 2x - 1$

${x}^{2} = - 1$

There is no real number whose square is -1.

Therefore, the equation has no solution.

6 0

3 years ago

Read 2 more answers

What is 34/38 in simplest form? ( please show the work)

olga55 [171]

34 and 38 are both multiples of 34 and 38 so divide each by 2
34/38 = 17/19
this is its simplest form as it cannot be cancelled down any more

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3 years ago

Let E = {(x, y) e R2|xy > 0}. Determine whether E is a subspace of R2 . X3

gayaneshka [121]

Answer:

E is not a subspace of $\mathbb{R}^2$

Step-by-step explanation:

E is not a subspace of $\mathbb{R}^2$

In order to see this, we must find two points (a,b), (c,d) in E such that (a,b) + (c,d) is not in E.

Consider

(a,b) = (1,1)

(c,d) = (-1,-1)

It is easy to see that both (a,b) and (c,d) are in E since 1*1>0 and (1-)*(-1)>0.

But (a,b) + (c,d) = (1-1, 1-1) = (0,0)

and (0,0) is not in E.

By the way, it can be proved that in any vector space all sub spaces must have the vector zero.

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3 years ago