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2 years ago

Can you please help me

Mathematics

1 answer:

Pepsi [2]2 years ago

6 0

Answer:

I can see the picture can u resend it?

Step-by-step explanation:

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A carpenter builds tables and chairs of various sizes and sells them to local businesses. She created the following chart for he

sukhopar [10]

The system of equations that can be used to determine the number of small tables and the number of large tables the designer will make is:

y -x = 8

$12y - $6x = $236

<h3>Defining the variables </h3>

Let:

x = number of large tables the designer can make.

x = number of small tables the designer can make.

<h3>Simultaneous equations</h3>

Simultaneous equations are sets of equations that must be solved together in order to determine the required variables. Methods of solving simultaneous equations include:

Elimination method
substitution method
Graph method

To learn more about simultaneous equations, please check: brainly.com/question/25875552

5 0

1 year ago

(3x-7)+(4x-8) what is the sum of this question pls

Papessa [141]

Answer:

17x-15 ................... I needed to put dots so I can have 20 characters

8 0

2 years ago

Read 2 more answers

RADIANS?.................

Taya2010 [7]

Answer:

pi/18

Step-by-step explanation:

To change from degrees to radians, we multiply by pi/180

10 * pi/180

pi/18

8 0

2 years ago

The sum of <br> 19<br> and Goran's age is <br> 62<br> .

sergey [27]

43 is the answer 62-19= 43

6 0

2 years ago

Expand (2x+2)^6<br> How would you find the answer using the binomial theorem?

Yanka [14]

Answer:

Step-by-step explanation:

$\displaystyle\\\sum\limits _{k=0}^n\frac{n!}{k!*(n-k)!}a^{n-k}b^k .\\\\k=0\\\frac{n!}{0!*(n-0)!}a^{n-0}b^0=C_n^0a^n*1=C_n^0a^n.\\\\ k=1\\\frac{n!}{1!*(n-1)!} a^{n-1}b^1=C_n^1a^{n-1}b^1.\\\\k=2\\\frac{n!}{2!*(n-2)!} a^{n-2}b^2=C_n^2a^{n-2}b^2.\\\\k=n\\\frac{n!}{n!*(n-n)!} a^{n-n}b^n=C_n^na^0b^n=C_n^nb^n.\\\\C_n^0a^n+C_n^1a^{n-1}b^1+C_n^2a^{n-2}b^2+...+C_n^nb^n=(a+b)^n.$

$\displaystyle\\(2x+2)^6=\frac{6!}{(6-0)!*0!} (2x)^62^0+\frac{6!}{(6-1)!*1!} (2x)^{6-1}2^1+\frac{6!}{(6-2)!*2!}(2x)^{6-2}2^2+\\\\ +\frac{6!}{(6-3)!*3!} (2a)^{6-3}2^3+\frac{6!}{(6-4)*4!} (2x)^{6-4}b^4+\frac{6!}{(6-5)!*5!}(2x)^{6-5} b^5+\frac{6!}{(6-6)!*6!}(2x)^{6-6}b^6. \\\\$

$(2x+2)^6=\frac{6!}{6!*1} 2^6*x^6*1+\frac{5!*6}{5!*1}2^5*x^5*2+\\\\+\frac{4!*5*6}{4!*1*2}2^4*x^4*2^2+ \frac{3!*4*5*6}{3!*1*2*3} 2^3*x^3*2^3+\frac{4!*5*6}{2!*4!}2^2*x^2*2^4+\\\\+\frac{5!*6}{1!*5!} 2^1*x^1*2^5+\frac{6!}{0!*6!} x^02^6\\\\(2x+2)^6=64x^6+384x^5+960x^4+1280x^3+960x^2+384x+64.$

8 0

1 year ago