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3 years ago

Diallo says that the value of 12-3 to the second power is 81. how did he get that answer

2 answers:

6 0

He would have to subtract 12 and 3 which would give you a total of 9. then he would have to find the value of 9 squared which is 81. Hope that helped!

notka56 [123]3 years ago

5 0

Answer:

12-3= 9

9^2= 81

Step-by-step explanation:

As the statement indicates that diallo says that the value of 12-3 to the second power is 81, it means that he subtracted 3 from 12 which is equal to 9. Also, the value of 12-3 to the second power means that he calculed the value of 9 squared which is 81.

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Answer:

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Step-by-step explanation:

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3 years ago

How manys days makes a leap year?

dem82 [27]

Answer: In an ordinary year, if you were to count all the days in a calendar from January to December, you'd count 365 days. But approximately every four years, February has 29 days instead of 28. So, there are 366 days in the year. This is called a leap year.

Step-by-step explanation:

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3 years ago

1) Determine the discriminant of the 2nd degree equation below:

Aleksandr-060686 [28]

$\LARGE{ \boxed{ \mathbb{ \color{purple}{SOLUTION:}}}}$

We have, Discriminant formula for finding roots:

$\large{ \boxed{ \rm{x = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac} }{2a} }}}$

Here,

x is the root of the equation.
a is the coefficient of x^2
b is the coefficient of x
c is the constant term

1) Given,

3x^2 - 2x - 1

Finding the discriminant,

➝ D = b^2 - 4ac

➝ D = (-2)^2 - 4 × 3 × (-1)

➝ D = 4 - (-12)

➝ D = 4 + 12

➝ D = 16

2) Solving by using Bhaskar formula,

❒ p(x) = x^2 + 5x + 6 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5\pm \sqrt{( - 5) {}^{2} - 4 \times 1 \times 6 }} {2 \times 1}}}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5 \pm \sqrt{25 - 24} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5 \pm 1}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = - 2 \: or - 3}}}$

❒ p(x) = x^2 + 2x + 1 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm \sqrt{ {2}^{2} - 4 \times 1 \times 1} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm \sqrt{4 - 4} }{2} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm 0}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = - 1 \: or \: - 1}}}$

❒ p(x) = x^2 - x - 20 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - ( - 1) \pm \sqrt{( - 1) {}^{2} - 4 \times 1 \times ( - 20) } }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ 1 \pm \sqrt{1 + 80} }{2} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{1 \pm 9}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = 5 \: or \: - 4}}}$

❒ p(x) = x^2 - 3x - 4 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - ( - 3) \pm \sqrt{( - 3) {}^{2} - 4 \times 1 \times ( - 4) } }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{3 \pm \sqrt{9 + 16} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{3 \pm 5}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = 4 \: or \: - 1}}}$

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5 0

3 years ago

Read 2 more answers

If a certain number of Grade IV students share 28 or 39 skill books in Mathematics,there are remainders of 4 and 3 books. What i

nadya68 [22]

Answer:

12 students

Step-by-step explanation:

Below numbers are divisible by the number of students:

28 - 4 = 24
39 - 3 = 36

Lets find GCF of 24 and 36:

24 = 2*2*2*3
36 = 2*2*3*3
GCF(24,36) = 2*2*3 = 12

The largest possible number of students is 12

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3 years ago

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umka21 [38]

Answer:

4 number

Step-by-step explanation:

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3 years ago