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3 years ago

9

Whoever gets this correct gets brain, 5 stars and a thanks! An explanation of it would be amazing! Tysm!

1 answer:

Jet001 [13]3 years ago

7 0

So the correct ratio is 4:5
The only one that doesn’t mean the same as that is C.

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What is the answer for 7÷ 4/5

Strike441 [17]

Answer:

7 ÷ 4/5 is 8.75 as a decimal or 8 3/4 as a fraction

7 0

2 years ago

Read 2 more answers

An individual is planning a trip to a baseball game for 16 people. Of the people planning to go to the baseball game, 8 can go o

Sedaia [141]

Answer: There are 4 people who only go to the game on Saturday.

Step-by-step explanation:

Let the number of people go on Saturday be n(A).

Let the number of people go on Sunday be n(B).

Since we have given that

n(A) = 8

n(B) = 12

n(A∪B) = 16

According to rules, we get that

$n(A)+n(B)-n(A\cap B)=n(A\cup B)\\\\8+12-n(A\cap B)=16\\\\20-n(A\cap B)=16\\\\n(A\cap B)=20-16=4$

So, n(only go on Saturday) = n(only A) = n(A) - n(A∩B) = 8-4 = 4

Hence, there are 4 people who only go to the game on Saturday.

4 0

3 years ago

6. If 2P and 2Q are complementary angles, m<P = 31º. What is mzQ?

Nat2105 [25]

Answer:

c

Step-by-step explanation:

5 0

2 years ago

(x +y)^5<br> Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=x,\:\:b=y$

$=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$

so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

$=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5$

$=1\cdot \:1\cdot \:x^5$

$=x^5$

also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5x^4y}{1}$

$=\frac{5x^4y}{1}$

$=5x^4y$

similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

$=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

Therefore,

$\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

6 0

2 years ago

Im an hour late, once again.. but i got the points<br><br> remember 10am and 3pm est

Iteru [2.4K]

Answer:

uh ok :

Step-by-step explanation:

3 0

3 years ago