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zhannawk [14.2K]
3 years ago
11

Zeina has seven times as many books as ali if zeina gave ali 15 books they would eash have the same number of books how many boo

ks did each have
Mathematics
1 answer:
ELEN [110]3 years ago
5 0

Step-by-step explanation:

20 each because it's like tgat

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A= P(1+nr) ; solve for r
Ugo [173]
The answer is r = -1/n+A/n
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3 years ago
(N + 1 )^ 2 for N = -9
Mariulka [41]

Answer: 82

Step-by-step explanation:

(N + 1 )^ 2 for N = -9

(-9 +1))^ 2

81+1

82

(-9 +1))^ 2 - Every number raised to power 2 positive your result will be positive

3 0
3 years ago
HELP ME PLEASE!!!!
Rama09 [41]
The answer is B   
√240= √16 * √15  (√16= 4)

answer is 4√15
5 0
4 years ago
Read 2 more answers
Solve the following proportion for the unknown variable: 15/d = 24/40
Pavel [41]
<span><u><em>Answer:</em></u>
d = 25

<u><em>Explanation:</em></u>
To get the value of d, we will need to isolate the d on one side of the equation.

<u>This can be done as follows:</u>
</span>\frac{15}{d} =  \frac{24}{40}<span>

<u>First, we will do cross multiplication as follows:</u>
40*15 = 24*d
600 = 24d

<u>Then, we will isolate the d as follows:</u>
24d = 600
</span>\frac{24d}{24} =  \frac{600}{24}
<span>
d = 25

Hope this helps :)</span>
4 0
4 years ago
Read 2 more answers
Special Triangles
Ugo [173]

Answer:

see explanation

Step-by-step explanation:

Using the cosine and tangent trigonometric ratios and the exact values

cos30° = \frac{\sqrt{3} }{2} and tan30° = \frac{1}{\sqrt{3} } , then

cos30° = \frac{adjacent}{hypotenuse} = \frac{6}{m} = \frac{\sqrt{3} }{2} ( cross- multiply )

\sqrt{3} × m = 12 ( divide both sides by \sqrt{3} )

m = \frac{12}{\sqrt{3} } ← rationalise by multiplying numerator/ denominator by \sqrt{3} )

m = \frac{12}{\sqrt{3} } × \frac{\sqrt{3} }{\sqrt{3} } = \frac{12\sqrt{3} }{3} = 4\sqrt{3}

-------------------------------------------------------------------------------

tan30° = \frac{opposite}{adjacent} = \frac{n}{6} = \frac{1}{\sqrt{3} } ( cross- multiply )

\sqrt{3} × n = 6 ( divide both sides by \sqrt{3} )

n = \frac{6}{\sqrt{3} } ← rationalise the denominator

n = \frac{6}{\sqrt{3} } × \frac{\sqrt{3} }{\sqrt{3} } = \frac{6\sqrt{3} }{3} = 2\sqrt{3}

-----------------------------------------------------------------------------------

7 0
3 years ago
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