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2 years ago

15

Leroy baked 63 cookies with 9 scoops of flour. How many scoops of flour does Leroy need in order to bake 70 cookies? Solve using

unit rates.

1 answer:

Marysya12 [62]2 years ago

6 0

Answer:

10.

Step-by-step explanation:

63 cookies = 9 scoops of flour.

<u>If we go up once with another scoop of flour,</u>

<u>you should get this:</u>

70 cookies = 10 scoops of flour.

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Factorise 9w² - 100<br><br><br>

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Answer:

$9\, w^{2} - 100 = (3\, w - 10) \, (3\, w + 10)$ .

Step-by-step explanation:

Fact:

$\begin{aligned} & (a - b)\, (a + b)\\ =\; & a^{2} + a\, b - a\, b - b^{2} \\ =\; & a^{2} - b^{2} \end{aligned}$ .

In other words, $(a^{2} - b^{2})$ , the difference of two squares in the form $a^{2}$ and $b^{2}$ , could be factorized into $(a - b)\, (a + b)$ .

In this question, the expression $(9\, w^{2} - 100)$ is the difference between two terms: $9\, w^{2}$ and $100$ .

$9\, w^{2}$ is the square of $3\, w$ . That is: $(3\, w)^{2} = 9\, w^{2}$ .
On the other hand, $10^{2} = 100$ .

Hence:

$9\, w^{2} - 100 = (3\, w)^{2} - (10)^{2}$ .

Apply the fact that $a^{2} - b^{2} = (a - b) \, (a + b)$ to factorize this expression. (In this case, $a = 3\, w$ whereas $b = 10$ .)

$\begin{aligned}& 9\, w^{2} - 100 \\ =\; & (3\, w)^{2} - (10)^{2} \\ = \; & (3\, w - 10)\, (3\, w + 10)\end{aligned}$ .

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The straight line, Line L has an equation of 4x+y=7. Find the equation of the straight line perpendicular to Line L that passes

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Answer:

Step-by-step explanation:

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perp. 1/4

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