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tatyana61 [14]
3 years ago
11

Rectangle poster is 3 times as long as wide perimeter is 24 in

Mathematics
1 answer:
Alexxandr [17]3 years ago
3 0

w= width

l = 3w

P = 2(l+w)

substitute

24 = 2(3w+w)

24 = 2(4w)

24 =8w

3 =w

The width is 3 inches

the length is 9 inches

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Ganezh [65]

Answer:

They are compatible

Step-by-step explanation:

The first thing is to say that an "ace" and that it is a "coarse"

"ace" is card number 1. Group A

"coarse" is a type of the deck, found from number 1 to card 13. Group B

Thus:

 Calculate A U B:

1 to 13 + 1 of the other types of cards in the deck.

At intersection B:

1 of "coarse"

Therefore, if group A is compatible with group B

7 0
3 years ago
What 12+9 <br> A 99 <br> B 4<br> C 0
chubhunter [2.5K]

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5 0
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<img src="https://tex.z-dn.net/?f=%5Cfrac%7B3%2F7%7D%7B7%5Csqrt%7B10%7D%2F2%20%7D" id="TexFormula1" title="\frac{3/7}{7\sqrt{10}
VashaNatasha [74]

Multiply the numerator and denominator by 7×2 = 14 to eliminate the denominators of those fractions:

\dfrac{\dfrac37}{\dfrac{7\sqrt{10}}2}\times\dfrac{14}{14}=\dfrac{3\times2}{7\sqrt{10}\times7}=\dfrac6{49\sqrt{10}}

Rationalize the denominator by multiplying both numerator and denominator by √10:

\dfrac6{49\sqrt{10}}\times\dfrac{\sqrt{10}}{\sqrt{10}}=\dfrac{6\sqrt{10}}{49(\sqrt{10})^2}=\dfrac{6\sqrt{10}}{49\times10}=\dfrac{6\sqrt{10}}{490}

Lastly, cancel the common factor of 2 in both the numerator and denominator (which comes from 6 = 2×3 and 490 = 2×245):

\dfrac{6\sqrt{10}}{490}=\dfrac{3\sqrt{10}}{245}}

8 0
3 years ago
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Anna [14]

Answer:

the ones with 16 dollar jeans

8 0
3 years ago
I need help on this question plz
drek231 [11]

Given is triangle with base = 10 cm and perpendicular height = 3 cm.

We know the formula for area of triangle is given by :-

Area = \frac{1}{2} × (Base) × (perpendicular height)

Area = \frac{1}{2} × (10) × (3)

Area = \frac{30}{2}

Area = 15 cm²

So, final answer is 15 cm².

3 0
3 years ago
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