What is 100 divided by the divine right of 10 to the power of 7

If a + b = -5 and x + y = -5, what is 6y + 6x - b - a

Novay_Z [31]

6(x +y) -(a +b) = 6(-5) -(-5)
= -25

5 0

3 years ago

Elizabeth’s aunt started a college savings account for her with $6,000. What is the balance of the account after 5 years if the

vaieri [72.5K]

Answer:

6750

Step-by-step explanation:

simple interest = P×r%×t

P = $6000

r = 2.5%

t = 5 years

I = 6000 × 2.5/100 × 5

= 750

6000 + 750 = $6750

7 0

3 years ago

Louisa’s savings change by -$6 each time she goes bowling. In all, it changed by -$66 during the summer. How many times did she

Sav [38]

Answer: She went $11$ times during the summer.

Step-by-step explanation:

If Louisa's savings change by $-6$ each time she goes bowling and it went to $-66$ , that means that Louisa went another $11$ times.

$\frac{1}{11}$

So therefore your answer would be 11.

6 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

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

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

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

$=5x^4y$

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

3 years ago

Find QN. Please show all the work on how you got your answer

Effectus [21]

QN = 28

Solution:

Given MNPQ is a parallelogram.

QT = 4x + 6 and TN = 5x + 4

To find the length of QN:

Let us solve it using the property of parallelogram.

Property of Parallelogram:

Diagonals of the parallelogram bisect each other.

Therefore, QT = TN

⇒ 4x + 6 = 5x + 4

Arrange like terms together.

⇒ 6 – 4 = 5x – 4x

⇒ 2 = x

⇒ x = 2

Substitute x = 2 in QT and TN

QT = 4(2) + 6 = 14

TN = 5(2) + 4 = 14

QN = QT + TN

= 14 + 14

QN = 28

The length of QN is 28.

3 0

3 years ago