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ruslelena [56]
2 years ago
14

Write 2-3 sentences describing your biggest takeaways from this activity. Use as much academic language (e.g., transversal, para

llel, congruent) as you can. (activity was lines, transversals, and angles)
Mathematics
1 answer:
Paul [167]2 years ago
6 0

Answer:

Evaluate the following exponential function when x = 1.

f (x) = 9 (12)Squared X + 12

F(1) =Evaluate the following exponential function when x = 1.

f (x) = 9 (12)Squared X + 12

F(1) =Evaluate the following exponential function when x = 1.

f (x) = 9 (12)Squared X + 12

F(1) =

Step-by-step explanation:Evaluate the following exponential function when x = 1.

f (x) = 9 (12)Squared X + 12 gg

F(1) = 50

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yawa3891 [41]
The identity property. the same applies for n*1=n.
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3 years ago
A local carpet company has been hired to carpet a planetarium which is in the shape of a circle. If the radius of the planetariu
lapo4ka [179]
This is the concept of application of geometry and financial mathematics; To calculate for the total cost of the carpet we proceed as follows;
Total cost=[area of the carpet]*[price per square yard]
area of carpet is given by:
Area=πr^2
where;
radius,r=6 yeards
=π*6^2
=113.1 square yards
therefore the price of the carpet will be:
113.1*14
=$1,583.4
5 0
3 years ago
PLEASE HELP ME!!! QUICKLY!!
podryga [215]

Answer:

4

Step-by-step explanation:

2L=w

32=LxW

so you put the 2L as the w

which is 32=Lx2L

32=2L^2

16=L^2

4=L

*To find the width you input 4 as L*

So 8=W

The shortest side would be 4

3 0
1 year 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
∆ABC is reflected about the line y = -x to give ∆A'B'C' with vertices A'(-1, 1), B'(-2, -1), C(-1, 0). What are the vertices of
blsea [12.9K]
Hi there!
Reflections across the line y = -x always go by the rule (-y, -x). We can use this rule to get our answer here. We are given the aftermath of the reflection coordinates, which are <span>A'(-1, 1), B'(-2, -1), and C'(-1, 0). All we have to do now is switch up the coordinate values and multiply them by -1. Here is the work - 
A'(-1, 1) => (1, -1) => x -1 => A(-1, 1)
B'(-2, -1) => (-1, -2) => x -1 => B(1, 2)
C'(-1, 0) => (0, -1) => x -1 => C(0, 1)
Therefore, the coordinates of Triangle ABC are A(-1, 1); B(1, 2); C(0,1). Hope this helped and have a phenomenal day!</span>
6 0
3 years ago
Read 2 more answers
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