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11111nata11111 [884]
3 years ago
11

Match each pair of monomials with their greatest common factor. Tiles 4p4q5 and 8p3q2 24pq6 and 16p3q5 8p3q6 and 4p2q5 16p3q2 an

d 24p2q 12p3q and 4p2q5 Pairs 4p2q5 arrowBoth 4p3q2 arrowBoth 8p2q arrowBoth 8pq5 arrowBoth
Mathematics
1 answer:
pashok25 [27]3 years ago
8 0
The GCF of the first two is 4p³q².The GCF of the second two is 8pq⁵.The GCF of the third two is 4p²q⁵.The GCF of the fourth two is 8p²q.The GCF of the fifth two is 4p²q.
To find the GCF of each pair, find the greatest number that will divide into each coefficient.  As for the variable portions, choose the variable that has the smallest exponent from each pair.

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Daniel has a set off green,blue, and red the red marblesmmake up exactly 1halfs of the set the set has 2 blue marble the number
Blizzard [7]

Answer:

There are <u>12 marbles</u> in Daniel's set.

Step-by-step explanation:

Given:

Daniel has a set off green,blue and red.

The red marbles make up exactly half of the set.

The set has 2 blue marbles.

The green marbles in set are as twice of the blue marbles in set.

Now, to find the number of marbles in Daniel's set.

Let the number of marbles in Daniel set be x.

So, the red marbles =  \frac{1}{2}\ of\  x =\frac{x}{2}

Blue marbles = 2.

And the green marbles = 2\times 2 =4.

According to question:

x=\frac{x}{2} +2+4

x=\frac{x}{2} +6

x=\frac{x+12}{2}

<em>Multiplying both sides by 2 we get:</em>

<em />2x=x+12<em />

<em>Subtracting both sides by </em>x<em> we get:</em>

x=12.

Therefore, there are 12 marbles in Daniel's set.

7 0
4 years ago
At the beginning of year 1, Bode invests $250 at an annual simple interest rate of 3%. He makes no deposits to or withdrawals fr
Dennis_Churaev [7]
The initial investment = $250
<span>annual simple interest rate of 3% = 0.03
</span>
Let the number of years = n
the annual increase = 0.03 * 250
At the beginning of year 1 ⇒ n = 1 ⇒⇒⇒ A(1) = 250 + 0 * 250 * 0.03 = 250

At the beginning of year 2 ⇒ n = 2 ⇒⇒⇒ A(2) = 250 + 1 * 250 * 0.03
At the beginning of year 3 ⇒ n = 3 ⇒⇒⇒ A(2) = 250 + 2 * 250 * 0.03
and so on .......
∴ <span>The formula that can be used to find the account’s balance at the beginning of year n is:
</span>
A(n) = 250 + (n-1)(0.03 • 250)
<span>At the beginning of year 14 ⇒ n = 14 ⇒ substitute with n at A(n)</span>
∴ A(14) = 250 + (14-1)(0.03*250) = 347.5

So, the correct option is <span>D.A(n) = 250 + (n – 1)(0.03 • 250); $347.50 </span>
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A solid figure with...<span>two circular bases and one curved surfaceis called a cylinder

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