Find the midpoint of A(5,8) and B(-9,-6)

1. Suppose y varies directly as x. If y = 3 when x = 15. then find x when y = 5.

KATRIN_1 [288]

Answer:

when y=3,X=15

here X is five times the value of y

so now when y= 5 then the value of X will be 25

hope you understand......

8 0

3 years ago

Can someone help? 25 points

gregori [183]

Answer:

t ≥ 18

Step-by-step explanation:

He has run 12 miles plus some more miles t. He need to run at least 30 miles

12+t ≥ 30

Subtract 12 from each side

12+t-12 ≥ 30-12

t ≥ 18

3 0

2 years ago

Read 2 more answers

BRAINLIEST ANSWER UP FOR GRABS!!

Svetach [21]

X = sqrt(5^2 - 3^2) = 4
y = sqrt(7^2 - 3^2) = sqrt 40

Horizontal distance = x + y + 3 = 7 + sqrt40

8 0

2 years ago

Read 2 more answers

Use prime factorization to find the LCM of 6, 25, and 30

RSB [31]

150 is the LCM of 6, 25, and 30.

8 0

3 years ago

WILL MARK BRAINLIEST!!! 40 POINTS!! ACTUAL ANSWERS, PLZZZ

steposvetlana [31]

Answer:

Part A:

$\left(x + 7\right)^{5}=x^{5} + 35 x^{4} + 490 x^{3} + 3430 x^{2} + 12005 x + 16807$

Part B:

The closure property describes cases when mathematical operations are CLOSED. It means that if you apply certain mathematical operations in a polynomial it will still be a polynomial. Polynomials are closed for sum, subtraction, and multiplication.

It means:

$\text{Sum of polynomials } \Rightarrow \text{ It will always be a polynomial}$

$\text{Subtraction of polynomials } \Rightarrow \text{ It will always be a polynomial}$

$\text{Multiplication of polynomials } \Rightarrow \text{ It will always be a polynomial}$

But when it is about division:

$\text{Division of polynomials } \Rightarrow \text{ It will not always/sometimes be a polynomial}$

Example of subtraction of polynomials:

$(2x^2+2x+3) - (x^2+5x+2)$

$x^2-3x+1$

Step-by-step explanation:

First, it is very important to define what is a polynomial in standard form:

It is when the terms are ordered from the highest degree to the lowest degree.

Therefore I can give:

$x^5-5x^4+3x^3-3x^2+7x+20$

but,

$x^5+3x^3-3x^2+7x+20-5x^4$ is not in standard form.

For this question, I can simply give the answer: $x^5-5x^4+3x^3-3x^2+7x+20$ and it is correct.

But I will create a fifth-degree polynomial using this formula

$(a+b)^n=\sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$

Also, note that

$\binom{n}{k}=\frac{n!}{(n-k)!k!}$

For $a=x \text{ and } b=7$

$\left(x + 7\right)^{5}=\sum_{k=0}^{5} \binom{5}{k} \left(x\right)^{5-k} \left(7\right)^k$

$\text{Solving for } k \text{ values: } 0, 1, 2, 3, 4 \text{ and } 5$

Sorry but I will not type every step for each value of $k$

The first one is enough.

For $k=0$

$\binom{5}{0} \left(x\right)^{5-0} \left(7\right)^{0}=\frac{5!}{(5-0)! 0!}\left(x\right)^{5} \left(7\right)^{0}=\frac{5!}{5!} \cdot x^5= x^{5}$

Doing that for $k$ values:

$\left(x + 7\right)^{5}=x^{5} + 35 x^{4} + 490 x^{3} + 3430 x^{2} + 12005 x + 16807$

8 0

2 years ago