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xz_007 [3.2K]
3 years ago
11

Please I need help: Find a number greater than 1 ​and​ less than 1000 that ​is both​ a perfect square and a perfect cube.

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

Answer:

64

Step-by-step explanation:

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State the missing reasons in this flow proof
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is there a photo for this problem? I solved what i coulda. givenb. givenc. def. of supplementary angles (i think)d.photo to support this?

e.

5 0
4 years ago
The mean of the weight is 19.6 lb with a standard deviation of 1.3 lb. Which weight is greater than 70% of the data?
salantis [7]

Answer:what’s the answer ???


Step-by-step explanation:


4 0
3 years ago
What are the zeroes of y=x^2+14x+40
a_sh-v [17]
y=x^2+14x+40 

y-x^2-14x-40=0 

In general, given a{x}^{2}+bx+c=0, there exists two solutions where

x=  \dfrac{14+2 \sqrt{y+9} }{-2} , \dfrac{14-2 \sqrt{y+9} }{-2} 

x=-7- \sqrt{y+9},-7+ \sqrt{y+9}
3 0
3 years ago
Find sin2a if sina=3/5 and a is in quadrant 2
LenKa [72]
Sin(a)= \frac{3}{5}

To find sin(2a) = 2*sin(a)*cos(a), we ned to find cos(a)

We know that,

sin²(a) + cos²(a) = 1

cos(a) = \sqrt{( 1- sin^2 a)}

cos(a) = \sqrt{1- ( \frac{3}{5} )^2} =  \sqrt{1-  \frac{9}{25}} =  \sqrt{ \frac{16}{25}} =  \frac{4}{5}

Thus, cos(a) = +- \frac{4}{5}

We know a is in second quadrant and thus cos(a) is -\frac{4}{5} because cosine is always negative in the second quadrant.

Therefore,

sin(2a) = 2*sin(a)*cos(a) = 2* \frac{3}{5} * \frac{-4}{5} =  \frac{-24}{25}


5 0
3 years ago
Giving brainlist for answering this question. Please be quick I am falling behind this question thanks.
Kisachek [45]

Answer:

15h + 8

15h+8 (no spaces)

Step-by-step explanation:

5h + 6 + 2 + 10h

5h + 10h + 6 + 2

15h + 8

Hope this helps!

7 0
3 years ago
Read 2 more answers
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