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2 years ago

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Sally has 6 boxes of colored pencils only three of the boxes contain 12 colored pencils each .the remaining boxes each contain 2

0 colored pencils.she packs the colored pencils into 8 packets.how many pencils are there in each packet??

1 answer:

balandron [24]2 years ago

4 0

There are 12 pencils in each packet. 12(3)*20(3)=96
96÷8=12

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The numerator and denominator of a fraction are in the ratio of 3 to 5. If the numerator and denominator are both increased by 2

ivolga24 [154]

Question:

The numerator and denominator of a fraction are in the ratio of 3 to 5. If the numerator and denominator are both decreased by 2, the fraction is now equal to $\frac{1}{2}$ .

If n = the numerator and d = the denominator, which of the following systems of equations could be used to solve the problem?

5n = 3d and n - 2 = 2d - 4

5n = 3d and 2n - 4 = d - 2

3n = 5d and 2n - 4 = d - 2

Answer:

5n = 3d and 2n – 4 = d – 2

Solution:

Let n be the numerator of the fraction and d be the denominator of the fraction.

Given the numerator and denominator of a fraction are in the ratio of 3 to 5.

This can be written as n : d = 3 : 5.

⇒ $\frac{n}{d}= \frac{3}{5}$ – – – – (1)

Do cross multiplication, we get

⇒ 5n = 3d

When the numerator and denominator are decreased by 2, the fraction is equal to $\frac{1}{2}$ .

⇒ $\frac{n-2}{d-2}= \frac{1}{2}$

Do cross multiplication, we get

⇒ 2(n –2)=1(d – 2)

⇒ 2n – 4 = d – 2

Hence, 5n = 3d and 2n – 4 = d – 2 can be used to solve the problem.

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3 years ago

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Which expression represents the number 2i4−5i3+3i2+−81‾‾‾‾√ rewritten in a+bi form?

vichka [17]

Answer:

The expression $-1+14i$ represents the number $2i^4-5i^3+3i^2+\sqrt{-81}$ rewritten in a+bi form.

Step-by-step explanation:

The value of $i$ is $i=\sqrt{-1}[tex] or [tex]i^{2}=-1[\tex].Now [tex]i^{4}$ in term of $i^{2}[\tex] can be written as, [tex]i^{4}=i^{2}\times i^{2}$

Substituting the value,

$i^{4}=\left(-1\right)\times \left(-1\right)$

Product of two negative numbers is always positive.

$\therefore i^{4}=1$

Now $i^{3}$ in term of $i^{2}[\tex] can be written as, [tex]i^{3}=i^{2}\times i$

Substituting the value,

$i^{3}=\left(-1\right)\times i$

Product of one negative and one positive numbers is always negative.

$\therefore i^{3}=-i$

Now $\sqrt{-81}$ can be written as follows,

$\sqrt{-81}=\sqrt{\left(81\right)\times\left(-1\right)}$

Applying radical multiplication rule,

$\sqrt{ab}={\sqrt{a}}\sqrt{b}$

$\sqrt{\left(81\right)\times\left(-1\right)}={\sqrt{81}}\sqrt{-1}$

Now, $\sqrt{\left(81\right)=9$ and $\sqrt{-1}}=i$

$\therefore \sqrt{\left(81\right)\times\left(-1\right)}=9i$

Now substituting the above values in given expression,

$2i^4-5i^3+3i^2+\sqrt{-81}=2\left(1\right)-5\left(-i\right)+3\left(-1\right)+9i$

Simplifying,

$2+5i-3+9i$

Collecting similar terms,

$2-3+5i+9i$

Combining similar terms,

$-1+14i$

The above expression is in the form of a+bi which is the required expression.

Hence, option number 4 is correct.

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