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Brut [27]
3 years ago
13

Nicole is making 1,000 bows for people who donate to the library book sale. She needs a piece of ribbon that is 0.75 meter long

for each bow, how many meters of ribbon does Nicole need to make the bows?
Mathematics
1 answer:
Natasha2012 [34]3 years ago
6 0
She needs 750 meters because 1,000 x .75 is 750 i thinkk
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Capital of Idaho is boise this is not a math question

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(x-2)=-1/4(x-8) <br> please show steps
RSB [31]

Answer:

(

−

2

)

=

−

1

4

(

−

8

)

(x-2)=\frac{-1}{4}(x-8)

(x−2)=4−1​(x−8)

Solve

1

Eliminate redundant parentheses

(

−

2

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4

(

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8

)

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2

=

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1

4

(

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8

)

2

Combine multiplied terms into a single fraction

−

2

=

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1

4

(

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8

)

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2

=

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1

(

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8

)

4

3

Distribute

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2

=

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1

(

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8

)

4

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2

=

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+

8

4

4

Add

2

2

2

to both sides of the equation

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2

=

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2

+

2

=

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5

Simplify

Add the numbers

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+

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4

+

2

Solution

=

1

6

5

Step-by-step explanation:

4 0
3 years ago
What is the mode of the histogram?
LUCKY_DIMON [66]

The mode is 7.

Mode means most and 7 hours of sleep is the most frequent.

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3 years ago
Solve the following question
White raven [17]

Answer:

g) u^{4}\cdot v^{-1}\cdot z^{3}, h) \frac{(x+4)\cdot (x+2)}{3\cdot (x-5)}

Step-by-step explanation:

We proceed to solve each equation by algebraic means:

g) \frac{u^{5}\cdot v}{z}\div  \frac{u\cdot v^{2}}{z^{4}}

1) \frac{u^{5}\cdot v}{z}\div  \frac{u\cdot v^{2}}{z^{4}} Given

2) \frac{\frac{u^{5}\cdot v}{z} }{\frac{u\cdot v^{2}}{z^{4}} } Definition of division

3) \frac{u^{5}\cdot v\cdot z^{4}}{u\cdot v^{2}\cdot z}   \frac{\frac{a}{b} }{\frac{c}{d} } = \frac{a\cdot d}{b\cdot c}

4) \left(\frac{u^{5}}{u} \right)\cdot \left(\frac{v}{v^{2}} \right)\cdot \left(\frac{z^{4}}{z} \right)  Associative property

5) u^{4}\cdot v^{-1}\cdot z^{3}   \frac{a^{m}}{a^{n}} = a^{m-n}/Result

h) \frac{x^{2}-16}{x^{2}-10\cdot x + 25} \div \frac{3\cdot x - 12}{x^{2}-3\cdot x -10}

1) \frac{x^{2}-16}{x^{2}-10\cdot x + 25} \div \frac{3\cdot x - 12}{x^{2}-3\cdot x -10} Given

2) \frac{\frac{x^{2}-16}{x^{2}-10\cdot x+25} }{\frac{3\cdot x - 12}{x^{2}-3\cdot x - 10} } Definition of division

3) \frac{(x^{2}-16)\cdot (x^{2}-3\cdot x -10)}{(x^{2}-10\cdot x + 25)\cdot (3\cdot x - 12)}  \frac{\frac{a}{b} }{\frac{c}{d} } = \frac{a\cdot d}{b\cdot c}

4) \frac{(x+4)\cdot (x-4)\cdot (x-5)\cdot (x+2)}{3\cdot (x-5)^{2}\cdot (x-4) } Factorization/Distributive property

5) \left(\frac{1}{3} \right)\cdot (x+4)\cdot (x+2)\cdot \left(\frac{x-4}{x-4} \right)\cdot \left[\frac{x-5}{(x-5)^{2}} \right] Modulative and commutative properties/Associative property

6) \frac{(x+4)\cdot (x+2)}{3\cdot (x-5)}  \frac{a^{m}}{a^{n}} = a^{m-n}/\frac{a}{b}\times \frac{c}{d} = \frac{a\cdot c}{b\cdot d}/Definition of division/Result

3 0
3 years ago
For integers a, b, and c, consider the linear Diophantine equation ax C by D c: Suppose integers x0 and y0 satisfy the equation;
Dmitrij [34]

Answer:

a.

x = x_1+r(\frac{b}{gcd(a, b)} )\\y=y_1-r(\frac{a}{gcd(a, b)} )

b. x = -8 and y = 4

Step-by-step explanation:

This question is incomplete. I will type the complete question below before giving my solution.

For integers a, b, c, consider the linear Diophantine equation

ax+by=c

Suppose integers x0 and yo satisfy the equation; that is,

ax_0+by_0 = c

what other values

x = x_0+h and y=y_0+k

also satisfy ax + by = c? Formulate a conjecture that answers this question.

Devise some numerical examples to ground your exploration. For example, 6(-3) + 15*2 = 12.

Can you find other integers x and y such that 6x + 15y = 12?

How many other pairs of integers x and y can you find ?

Can you find infinitely many other solutions?

From the Extended Euclidean Algorithm, given any integers a and b, integers s and t can be found such that

as+bt=gcd(a,b)

the numbers s and t are not unique, but you only need one pair. Once s and t are found, since we are assuming that gcd(a,b) divides c, there exists an integer k such that gcd(a,b)k = c.

Multiplying as + bt = gcd(a,b) through by k you get

a(sk) + b(tk) = gcd(a,b)k = c

So this gives one solution, with x = sk and y = tk.

Now assuming that ax1 + by1 = c is a solution, and ax + by = c is some other solution. Taking the difference between the two, we get

a(x_1-x) + b(y_1-y)=0

Therefore,

a(x_1-x) = b(y-y_1)

This means that a divides b(y−y1), and therefore a/gcd(a,b) divides y−y1. Hence,

y = y_1+r(\frac{a}{gcd(a, b)})  for some integer r. Substituting into the equation

a(x_1-x)=rb(\frac{a}{gcd(a, b)} )\\gcd(a, b)*a(x_1-x)=rba

or

x = x_1-r(\frac{b}{gcd(a, b)} )

Thus if ax1 + by1 = c is any solution, then all solutions are of the form

x = x_1+r(\frac{b}{gcd(a, b)} )\\y=y_1-r(\frac{a}{gcd(a, b)} )

In order to find all integer solutions to 6x + 15y = 12

we first use the Euclidean algorithm to find gcd(15,6); the parenthetical equation is how we will use this equality after we complete the computation.

15 = 6*2+3\\6=3*2+0

Therefore gcd(6,15) = 3. Since 3|12, the equation has integral solutions.

We then find a way of representing 3 as a linear combination of 6 and 15, using the Euclidean algorithm computation and the equalities, we have,

3 = 15-6*2

Because 4 multiplies 3 to give 12, we multiply by 4

12 = 15*4-6*8

So one solution is

x=-8 & y = 4

All other solutions will have the form

x=-8+\frac{15r}{3} = -8+5r\\y=4-\frac{6r}{3} =4-2r

where r ∈ Ζ

Hence by putting r values, we get many (x, y)

3 0
3 years ago
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