Find all the natural numbers, which when divided by 7, leave the same number for the quotient and the remainder.

Mathematics

1 answer:

Anna11 [10]4 years ago

5 0

Answer:

8,16,24,32,48

Step-by-step explanation:

When dividing a number by 7, only 7 different remainders are possible.

The remainders when dividing by 7 can not exceed 6, so the numbers in which we are interested can be represented as 7k+k or just 8k, where k=1, 2, ... , 6.

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The range of y = - 32x ^ 2 + 90x + 3

mariarad [96]

Given:

The given function is:

$y=-32x^2+90x+3$

To find:

The range of the given function.

Solution:

We have,

$y=-32x^2+90x+3$

It is a quadratic function because the highest power of the variable x is 2.

Here, the leading coefficient is -32 which is negative. So, the graph of the given function is a downward parabola.

If a quadratic function is $f(x)=ax^2+bx+c$ , then the vertex of the quadratic function is:

$Vertex=\left(\dfrac{-b}{2a},f(\dfrac{-b}{2a})\right)$

In the given function, $a=-32,b=90,c=3$ .

$\dfrac{-b}{2a}=\dfrac{-90}{2(-32)}$

$\dfrac{-b}{2a}=\dfrac{-90}{-64}$

$\dfrac{-b}{2a}=\dfrac{45}{32}$

The value of the given function at $x=\dfrac{45}{32}$ is:

$y=-32(\dfrac{45}{32})^2+90(\dfrac{45}{32})+3$

$y=\dfrac{2121}{32}$

The vertex of the given downward parabola is $\left(\dfrac{45}{32},\dfrac{2121}{32}\right)$ . It means the maximum value of the function is $y=\dfrac{2121}{32}$ . So,