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Bogdan [553]
3 years ago
8

True or false any nonzero integer divided by zero equals zero

Mathematics
2 answers:
Rzqust [24]3 years ago
6 0
This is true!!!!!!!!!!

Kaylis [27]3 years ago
4 0
That's false.  Division by zero isn't even allowed, because it's undefined.
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digit at tens place of two digit number is twice the digit at units place .if sum of this number and the no.formed by reversing
Dima020 [189]
Let the number be     xy ,

xy = 10x + y.   when reversed = yx = 10y + x

x = 2y          ........(a)

10x + y + 10y + x = 66.........(b)

10(2y) + y + 10y + 2y = 66

20y + y + 10y + 2y = 66

33y = 66

y = 66/33

y = 2

x = 2y = 2*2 = 4

Recall the original number is xy =     42.
8 0
3 years ago
What is the definition of evaluate? this is for a test please help​
Daniel [21]

Answer:

The definition of evaluate is to find a numerical expression or equivalent for - an equation, formula or function.

Step-by-step explanation:

6 0
3 years ago
PLEASE HELP!! 10 POINTS
kenny6666 [7]

Check the picture below.

we know B is the midpoint of AC, and we know ED is 4 units, since AD is a total of 8, then AE must also be 4 units, meaning point E is the midpoint of AD.

since E is the midpoint of AD and B is the midpoint of AC, then EB is the midsegment of the triangle, and is parallel to the base DC.

∡ABE = 60°

∡AEB = 90°

3 0
3 years ago
Find all the zeroes of the equation(with simple steps).
uysha [10]

<u>Answer-</u>

<em>The zeros are, 5,\ -5,\ 4i,\ -4i</em>

<u>Solution-</u>

\Rightarrow -3x^4+27x^2+1200=0

\Rightarrow -3(x^2)^2+27(x^2)+1200=0

Here,

a = -3, b = 27, c = 1200

So,

x^2=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}

=\dfrac{-27\pm \sqrt{-27^2-4\cdot (-3)\cdot 1200}}{2\cdot (-3)}

=\dfrac{-27\pm \sqrt{729+14400}}{-6}

=\dfrac{-27\pm 123}{-6}

=\dfrac{-27+123}{-6},\ \dfrac{-27- 123}{-6}

=\dfrac{96}{-6},\ \dfrac{-150}{-6}

=-16,\ 25

So,

\Rightarrow x^2=25,\ -16

\Rightarrow x=\sqrt{25},\ \sqrt{-16}

\Rightarrow x=5,\ -5,\ 4i,\ -4i

8 0
3 years ago
Anyone know the answer?
Alborosie

Step-by-step explanation:

{9}^{ \frac{ - 1}{2} }  \\  {3}^{2 \times  -  \frac{1}{2} }  \\  {3}^{ - 1}  \\  \frac{1}{3}

Now for another

{27}^{ \frac{ - 2}{3} }  \\  {3}^{3 \times  \frac{ - 2}{3} }  \\  {3}^{ - 2} \\   \frac{1}{9}

Hope it will help :)❤

7 0
3 years ago
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