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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?

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:

(

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2

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=

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4

(

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)

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

(x−2)=4−1(x−8)

Solve

1

Eliminate redundant parentheses

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=

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4

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2

Combine multiplied terms into a single fraction

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Distribute

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4

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Add

2

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to both sides of the equation

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Simplify

Add the numbers

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Solution

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6

5

Step-by-step explanation:

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The mode is 7.

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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