How do I solve this lol

Mathematics

1 answer:

Olin [163]3 years ago

7 0

I would plug the X values into the X spot in your equation and plug in your Y values into the Y spot and dove the equation from there. That pretty much all of the advice I can give for you didn’t provide the actual question but I hope the helped!

You might be interested in

According to a student survey, 16 students liked history, 19 liked English, 18 liked math, 8 liked math and English, 5 liked his

Bogdan [553]

When building a Venn Diagram, I always start from the area with the most overlap to the areas of least overlap. Once you have placed the 3 in the middle, you have counted those people, and therefore you must subtract them from the other surveys. Example: since there are 3 people that like all three subjects, now only have 5 students that like just math and English instead of 8.

Therefore:
A) 36 Students were in the survey
*Add all the numbers within the Venn diagram up. Overlapping doesn't matter because no one is double counted.

B) 6 People liked only Math
*Can't touch any other circle but Math

C) 20 Students liked English and math, but not history
*You add 9+5+6, since these bubbles are not overlapping with history.

I Hope this helps and let me know if you have any further questions!

4 0

3 years ago

This diagram shows a sector of a circle radius 11cm

denis-greek [22]

Step-by-step explanation:

first you have to use the formula for circle is 2πr

π= ----

and r stands for radius

5 0

1 year ago

What is the fourth term in the binomial expansion (a+b)^6)

Dafna11 [192]

Answer:

$20a^3b^3$

Step-by-step explanation:

<u>Binomial Series</u>

$(a+b)^n=a^n+\dfrac{n!}{1!(n-1)!}a^{n-1}b+\dfrac{n!}{2!(n-2)!}a^{n-2}b^2+...+\dfrac{n!}{r!(n-r)!}a^{n-r}b^r+...+b^n$

<u>Factorial</u> is denoted by an exclamation mark "!" placed after the number. It means to multiply all whole numbers from the given number down to 1.

Example: 4! = 4 × 3 × 2 × 1

Therefore, the fourth term in the binomial expansion (a + b)⁶ is:

$\implies \dfrac{n!}{3!(n-3)!}a^{n-3}b^3$

$\implies \dfrac{6!}{3!(6-3)!}a^{6-3}b^3$

$\implies \dfrac{6!}{3!3!}a^{3}b^3$

$\implies \left(\dfrac{6 \times 5 \times 4 \times \diagup\!\!\!\!3 \times \diagup\!\!\!\!2 \times \diagup\!\!\!\!1}{3 \times 2 \times 1 \times \diagup\!\!\!\!3 \times \diagup\!\!\!\!2 \times \diagup\!\!\!\!1}\right)a^{3}b^3$