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2 years ago

Boris chooses three DIFFERENT numbers.

Mathematics

1 answer:

g100num [7]2 years ago

6 0

Answer:

4, 5, 27.

Step-by-step explanation:

Let us find our numbers using given information.

Factors of 20 are: 1,2,4,5,10 and 20.

Let us try by taking number 4 and 5.

$36-(4+5)=36-9=27$

We can see that $3^{3} =27$

Now let us find sum of these numbers.

$4+5+27=36$

Therefore, numbers chosen by Boris will be 4, 5 and 27 as these numbers satisfies our all conditions.

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Which graph correctly solves the system of equations below? y = −x2 − 1 y = 2x2 − 4 (6 points)

myrzilka [38]

The equations are

$i) y=-x^2-1\\\\ii) y=2x^2-4$ ,

The graphs of the solutions (x, y) of these equations are 2 parabolas, since the right hand side expressions are polynomials of degree 2.

The solution/s of the system are the x-coordinates of the point/s of intersection of the parabolas.

The solutions of the first equation form a parabola looking downwards (since the coefficient of x^2 is -), and the second, a parabola opening upwards (since the coefficient of x^2 is +).

We can draw both parabolas, but to find the solution we still need to solve the system algebraically.

The algebraic solution of the system is:

$-x^2-1=2x^2-4\\\\3x^2-3=0\\\\3(x^2-1)=0\\\\x^2-1=0$ , so

the solutions are x=-1 and x=1.

The graph of the system is drawn using desmos.com

If we are allowed to use a graphic calculator, we can draw both graphs and point at the solution.

8 0

2 years ago

Read 2 more answers

What is the sum of the matrices shown below?

damaskus [11]

Answer:

$\left[\begin{array}{cc}3&9\\5&-2\end{array}\right] +\left[\begin{array}{cc}6&0\\-8&4\end{array}\right]=\left[\begin{array}{cc}9&9\\-3&2\end{array}\right]$

Step-by-step explanation:

To add matrices, we add the corresponding components.

The given matrices is

$\left[\begin{array}{cc}3&9\\5&-2\end{array}\right] +\left[\begin{array}{cc}6&0\\-8&4\end{array}\right]$

We add the corresponding components to get;

$\left[\begin{array}{cc}3&9\\5&-2\end{array}\right] +\left[\begin{array}{cc}6&0\\-8&4\end{array}\right]=\left[\begin{array}{cc}3+6&9+0\\5+-8&-2+4\end{array}\right]$

We simplify to get:

$\left[\begin{array}{cc}3&9\\5&-2\end{array}\right] +\left[\begin{array}{cc}6&0\\-8&4\end{array}\right]=\left[\begin{array}{cc}9&9\\-3&2\end{array}\right]$

5 0

2 years ago

Read 2 more answers

A school sold small and large boxes of fruit for a fundraiser. One student sold 5 small boxes and 13 large boxes for $232. Anoth

algol [13]

Answer: the right answer is

The first equation can be multiplied by –12 and the second equation by 5 to eliminate x.

The first equation can be multiplied by –7 and the second equation by 13 to eliminate y.

The first equation can be multiplied by 7 and the second equation by –13 to eliminate y.

The first equation can be multiplied by 12 and the second equation by –5 to eliminate x.

Step-by-step explanation:

e d g e n u i t y and it says check all that apply ,it was not (A).