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ch4aika [34]
4 years ago
13

What is the distance from home to school?

Mathematics
2 answers:
xxMikexx [17]4 years ago
8 0
The distance is 160ft!

liubo4ka [24]4 years ago
3 0

straight line from home to school =

sqrt(80^2 + 80^2) =

sqrt(12800) = 113.137 = 113.14 feet

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kirza4 [7]
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4.Now you have (7y-24).
Hoped this helps!!:-)))
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Proving lines parallel determine the value of X
Maslowich

<em>when </em><em>both </em><em>l </em><em>and </em><em>m </em><em>are </em><em>parallel</em><em> </em><em>then </em>

<em>(</em><em>4</em><em>x</em><em>+</em><em>7</em><em>)</em><em> </em><em> </em><em>=</em><em> </em><em>(</em><em>6</em><em>x</em><em>-</em><em>6</em><em>3</em><em>)</em>

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<em>-</em><em>x </em><em>=</em><em> </em><em>-</em><em>7</em><em>0</em><em>/</em><em>2</em>

<em>-x </em><em>=</em><em> </em><em>-</em><em>3</em><em>5</em>

<em>x </em><em>=</em><em> </em><em>3</em><em>5</em>

<em>hope </em><em>it </em><em>helps</em><em> </em><em>and </em><em>ur </em><em>day </em><em>will </em><em>full </em><em>of </em><em>happiness</em>

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3 years ago
50 POINTS
user100 [1]

\huge \sf༆ Answer ༄

According to the question ;

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:y \propto x

Now, let's add a constant of proportionality to replace proportionality sign with equal to :

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:y = kx

[ where, k is proportionality constant ]

plug the value of x and y from any ordered pair in the table , for example : \sf(\dfrac{1}{5} , 2)

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:2 = k \times  \dfrac{1}{5}

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:k = 2 \times 5

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:k = 10

Therefore, value of proportionality constant (k) is

\overbrace{  \underbrace{\underline{ \boxed{ \sf 10 }}}}

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Stuck on this questions
Viktor [21]

Answer:

1. connotative 2. actually kicking a bucket

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2. Denotative is literal so they are literally kicking a bucket.

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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}  

<u>Example of subtraction of polynomials:</u>

<u />(2x^2+2x+3) - (x^2+5x+2)<u />

<u />x^2-3x+1<u />

Step-by-step explanation:

<u>First, it is very important to define what is a polynomial in standard form: </u>

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
3 years ago
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