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Colt1911 [192]
2 years ago
10

What is the greatest perfect square that is a factor of 1290

Mathematics
2 answers:
AveGali [126]2 years ago
6 0

Answer:

There is no greatest perfect square that is a factor of 1290.

Step-by-step explanation:

The factors of 1290 are 1, 2, 3, 5, 6, 10, 15, 30, 43, 86, 129, 215, 258, 430, 645, 1290. There are no perfect squares as factors.

Keith_Richards [23]2 years ago
3 0

Answer:

There is no greatest perfect square of 1290

Step-by-step explanation:

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Here is a rectangle ABCD.
lana66690 [7]

Answer:

32%

Step-by-step explanation:

Area of rectangle ABCD = 20 * 30 = 600 Sq cm

After increasing the length and width of the rectangle by 20% and 10% respectively.

New length = 30 + 20% of 30 = 30 + 6 = 36 cm

New width = 20 + 10% of 20 = 20 + 2 = 22 cm

Area of new rectangle = 36*22 = 792 Sq cm

Increase in area = 792 - 600 = 192 Sq cm

Percentage increase in area

= (192/600) *100

= 32%

5 0
3 years ago
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Molly had a balance of $72 In her savings account before depositing $25. What is her new balance?
vovangra [49]

Answer:

$97

Step-by-step explanation:

just add 72 and 25 dude

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3 years ago
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ser-zykov [4K]

Answer:

The answer is C. -336

Step-by-step explanation:

All you have to do is 25 - 39 = -14 x 24 = -336

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3 years ago
~**Will mark brainliest **~<br> For the correct answers to all three questions
aleksandr82 [10.1K]

a)

\dfrac{42}{65}\cdot\dfrac{25}{36}\cdot\dfrac{26}{49}=\dfrac{2\cdot3\cdot7}{5\cdot13}\cdot\dfrac{5\cdot5}{2\cdot2\cdot3\cdot3}\cdot\dfrac{2\cdot13}{7\cdot7}=\dfrac{7}{13}\cdot\dfrac{5}{2\cdot3}\cdot\dfrac{2\cdot13}{7\cdot7}=\\\\\\=\dfrac{1}{1}\cdot\dfrac{5}{2\cdot3}\cdot\dfrac{2}{7}=\dfrac{5}{3}\cdot\dfrac{1}{7}=\dfrac{5}{21}

b)

\dfrac{21}{32}\cdot\dfrac{39}{120}\cdot\dfrac{40}{65}=\dfrac{21}{32}\cdot\dfrac{3\cdot13}{2\cdot2\cdot2\cdot3\cdot5}\cdot\dfrac{2\cdot2\cdot2\cdot5}{5\cdot13}=\\\\\\=\dfrac{3\cdot7}{32}\cdot\dfrac{13}{2\cdot2\cdot2\cdot5}\cdot\dfrac{2\cdot2\cdot2}{13}=\dfrac{21}{32}\cdot\dfrac{1}{5}\cdot\dfrac{1}{1}=\dfrac{21}{160}

c)

\dfrac{15}{90}\cdot\dfrac{36}{75}\cdot\dfrac{27}{42}=\dfrac{3\cdot5}{2\cdot3\cdot3\cdot5}\cdot\dfrac{2\cdot2\cdot3\cdot3}{3\cdot5\cdot5}\cdot\dfrac{3\cdot3\cdot3}{2\cdot3\cdot7}=\\\\\\=\dfrac{1}{2\cdot3}\cdot\dfrac{2\cdot2\cdot3}{5\cdot5}\cdot\dfrac{3\cdot3}{2\cdot7}=\dfrac{1}{1}\cdot\dfrac{2}{5\cdot5}\cdot\dfrac{3\cdot3}{2\cdot7}=\dfrac{1}{25}\cdot\dfrac{9}{7}=\dfrac{9}{175}

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3 years ago
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Phantasy [73]

Answer:

ΔABC≅ΔDEC by AAS

Step-by-step explanation:

You can use the AAS method of congruency.

Since you already have <BAC and <EDC congruent to eachother, and sides BC and EC congruent to each other, you only need that one remaining angle in between. <ACB can be proven congruent to <DCE by the Vertical Angles Theorem, and that gives you the AAS you need to prove that these two triangles are congruent

Hope this helped.

5 0
2 years ago
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