Maximize Y = 1.27 - 0.03x - 0.05y, using the constraints:

Factor the following.

Verizon [17]

The answer is D (y+10)(y-6)
because

6 0

3 years ago

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Which of this fraction are greater then1/2 but less then 3/4 .7/8 .2/5 .1/3 .5/8 .3/6

avanturin [10]

Answer: 7/8

Step-by-step explanation:

think of each faction as a pie or pizza.

each one will have the number of slices as the bottom number of the fraction.

now shade in the top numbers and see if it more than your 1/2

you will see 7/8 is the answer since it will be one piece left from being complete or "gone"

hope this helped angel!!

4 0

3 years ago

The producer of the news station posted an article about the high school's football championship ceremoney on a new website. The

-Dominant- [34]

Let x be the number of hours and Let y be the number of views.

x y
1 125
2 250
3 375
4 500
5 625

3 0

3 years ago

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Find the value of x?

Burka [1]

Answer:

We know that the exterior angle of a triangle is equal to the sum of two opposite interior angle. 

Step-by-step explanation:

So

25x + 57 + x = 45x

26x + 57 = 45x

57 = 45x - 26x

19x = 57

Therefore x = 3

Hope it will help you

3 0

3 years ago

(x +y)^5 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

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

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

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

$=5x^4y$

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