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3 years ago

6,8,5,8,8,8,1,8,1 mode: range:

Mathematics

2 answers:

iragen [17]3 years ago

8 0

Answer:

mode=8 because it is most repeated

range=7 because largest -smallest=8-1= 7,so..

Roman55 [17]3 years ago

3 0

Answer:

mode is 8

range is 7

Step-by-step explanation:

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(Identify the Terms and Like Terms in the expression) 10x+5+3x+1

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Please help, question is below

dimulka [17.4K]

The number of positive zeros is either 2 or 0. The number of negative zeros is either 2 or 0

<h3>How to determine the true statement?</h3>

The polynomial function is given as:

$f(x)=x^4-2x^3\ +\ 11x^2+5x\ +\ 6$

The degree of the polynomial function is:

Degree = 4

Divide by 2

Degree/2 = 2

This means that:

The number of positive zeros and negative zeros can be 2

The difference between these zeros count is:

Difference= 2 - 2 = 0

The number of positive zeros and negative zeros can also be 0

Hence, the number of positive zeros is either 2 or 0. The number of negative zeros is either 2 or 0

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2 years ago

Read 2 more answers

Which point is the solution to the following system of equations?

Naya [18.7K]

The point (3, 2) is the solution to given system of equations

Solution:

Given that system of equations are:

$x^2 + y^2 = 13$ ------ eqn 1

$2x - y = 4$ ------- eqn 2

From eqn 2,

y = 2x - 4

Substitute y = 2x - 4 in eqn 1

$x^2 + (2x - 4)^2 = 13\\\\x^2 + 4x^2 + 16 - 16x = 13\\\\5x^2 -16x + 3 = 0$

Let us solve the above equation by quadratic formula,

$\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0\\\\x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Using the Quadratic Formula for $5x^2 -16x + 3 = 0$ where a = 5, b = -16, and c = 3

$\begin{aligned}&x=\frac{-(-16) \pm \sqrt{(-16)^{2}-4(5)(3)}}{2 \times 5}\\\\&x=\frac{16 \pm \sqrt{256-60}}{10}\\\\&x=\frac{16 \pm \sqrt{196}}{10}\end{aligned}$

The discriminant $b^2 - 4ac>0$ so, there are two real roots.

$\begin{aligned}&x=\frac{16 \pm \sqrt{196}}{10}=\frac{16 \pm 14}{10}\\\\&x=\frac{16+14}{10} \text { or } \frac{16-14}{10}\\\\&x=\frac{30}{10} \text { or } x=\frac{2}{10}\\\\&x=3 \text { or } x=0.2\end{aligned}$

Substitute for x = 0.2 and x = 3 in 2x - y = 4

when x = 3

2(3) - y = 4

6 - y = 4

y = 2

when x = 0.2

2(0.2) - y = 4

0.4 - y = 4

y = 0.4 - 4

y = -3.6

Thus Option D is correct The point is (3, 2)

7 0

3 years ago