Which of the binomials below is a factor of this trinomial?

please help me answer this question Solve: y − x = 12 y + x = -26 (19, -7). (-7, 1). (7, 19). (-19, -7).

Tpy6a [65]

Answer:

(-19 , -7)

Step-by-step explanation:

y - x = 12

y + x = -25 we sum them to get

2y = -14 , y = -7

then we put -7 instead of y in any of the equations:

-7 - x = 12

-x = 19

x = -19,

finally (x , y) is (-19 , -7)

3 0

3 years ago

X to the second power = 500 what values can be used SQUARE ROOTS ONLY PLEASEEE!!!!!

Dahasolnce [82]

500
because when you power to the second power a square root, it cancels and the answer is the number you originally had

5 0

3 years ago

How do you write the circumference of the circle in terms on 几

Thepotemich [5.8K]

____________________________________

Hey!!!

The formula to find circumference of circle 

=2 pi r

For eg:

Radius=4 cm

Circumference of circle=2 pi r

 =2 *pi*4

 =8 pi

Hope it helps..

Good luck on your assignment

#Stay safe#

__________________________________

6 0

3 years ago

4 times a number is 16

andrey2020 [161]

Answer:

4 times 4 equals 16

Step-by-step explanation:

4 times 1= 4

4 times 2= 8

4 times 3= 12

4 times 4= 16

you add 4 after each result.

4 0

3 years ago

A = 1011 + 337 + 337/2 +1011/10 + 337/5 + ... + 1/2021

egoroff_w [7]

The sum of the given series can be found by simplification of the number

of terms in the series.

A is approximately 2020.022

Reasons:

The given sequence is presented as follows;

A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021

Therefore;

$\displaystyle A = \mathbf{1011 + \frac{1011}{3} + \frac{1011}{6} + \frac{1011}{10} + \frac{1011}{15} + ...+\frac{1}{2021}}$

The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;

$\displaystyle a_{n+1} = \mathbf{\frac{n^2 + 3 \cdot n + 2}{2}}$

Therefore, for the last term we have;

$\displaystyle 2043231= \frac{n^2 + 3 \cdot n + 2}{2}$

2 × 2043231 = n² + 3·n + 2

Which gives;

n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0

Which gives, the number of terms, n = 2020

$\displaystyle \frac{A}{2} = \mathbf{ 1011 \cdot \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460} \right)}$

$\displaystyle \frac{A}{2} = 1011 \cdot \left(1 - \frac{1}{2} +\frac{1}{2} - \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022} \right)$

Which gives;

$\displaystyle \frac{A}{2} = 1011 \cdot \left(1 - \frac{1}{2022} \right)$

$\displaystyle A = 2 \times 1011 \cdot \left(1 - \frac{1}{2022} \right) = \frac{1032231}{511} \approx \mathbf{2020.022}$

A ≈ 2020.022

Learn more about the sum of a series here:

brainly.com/question/190295

8 0

2 years ago

Read 2 more answers