All categories

3 years ago

Any sum one of this help plzz

Mathematics

2 answers:

neonofarm [45]3 years ago

8 0

What you do is plug in what x is equal to for the problem then use PEMDAS to figure the rest out

Digiron [165]3 years ago

7 0

Answer:

x=2

Step-by-step explanation: because the highest exponent is 2 then the value of the polynomial would be 2

You might be interested in

What are the x-intercepts of the quadratic function?

maw [93]

To find the x intercept set y = 0

x^2 + x - 2 =0

(x+2)(x-1) = 0

x = -2 , x = 1

8 0

2 years ago

Read 2 more answers

Give ur answer in standard form. <br>3 x 10^4 ÷6 x 10^-4<br>

Anni [7]

Answer:

5.0×10⁷

Step-by-step explanation:

(3÷6)×(10⁴÷10^-4)

(0.5)×(10⁸)

5.0×10^-1×10⁸

5.0×10⁷

7 0

1 year ago

Read 2 more answers

Which expression is equivalent to *picture attached*

DiKsa [7]

Answer:

The correct option is;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$

Step-by-step explanation:

The given expression is presented as follows;

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right )$

Which can be expanded into the following form;

$\sum\limits _{n = 1}^{50} \left (4\cdot n^2 + 3 \cdot n\right ) = 4 \times \sum\limits _{n = 1}^{50} \left n^2 + 3 \times\sum\limits _{n = 1}^{50} n$

From which we have;

$\sum\limits _{k = 1}^{n} \left k^2 = \dfrac{n \times (n+1) \times(2n+1)}{6}$

$\sum\limits _{k = 1}^{n} \left k = \dfrac{n \times (n+1) }{2}$

Therefore, substituting the value of n = 50 we have;

$\sum\limits _{n = 1}^{50} \left k^2 = \dfrac{50 \times (50+1) \times(2\cdot 50+1)}{6}$

$\sum\limits _{k = 1}^{50} \left k = \dfrac{50 \times (50+1) }{2}$

Which gives;

$4 \times \sum\limits _{n = 1}^{50} \left n^2 = 4 \times \dfrac{n \times (n+1) \times(2n+1)}{6} = 4 \times \dfrac{50 \times (50+1) \times(2 \times 50+1)}{6}$

$3 \times\sum\limits _{n = 1}^{50} n = 3 \times \dfrac{n \times (n+1) }{2} = 3 \times \dfrac{50 \times (51) }{2}$

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right ) = 4 \times \dfrac{50 \times (50+1) \times(2\times 50+1)}{6} +3 \times \dfrac{50 \times (51) }{2}$