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2 years ago

Find the area and perimeter

Mathematics

1 answer:

artcher [175]2 years ago

3 0

Answer:

Step-by-step explanation:

Add all 4 sides to get perimeter so the first one is 9+9+2+2=22cm

the area is length x width so the first one is 9 x 2 =18cm ^2

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12. Fancy goldfish x cost $3 each and

o-na [289]

Answer:

x (Fancy Gold fish) = 3

y (Common goldfish) = 29

Step-by-step explanation:

Fancy goldfish x cost $3 each

common goldfish y cost $1 each

Graph: y = 20 + 3x (I'm hoping this is the correct line equation, please do leave a comment below if it's not)

Thus, we graph it by subtitude either the x or y values. (Or you can use just X = 1,2,3).

Now, to find how many of each type of goldfish Tasha can buy. We subtitude the line equation by either using x = 3 or y = 1

Using either values to solve the equation, we get:

x = 3

y = 29

Hope this helps!

5 0

2 years ago

Read 2 more answers

If 2/5n+4=20, what is the value of n?

Dimas [21]

2/5n + 4 = 20
/5 /5
2n + 4 = 100
- 4 -4
2n = 96
/2 /2
n = 48

The answer to your question is 48

6 0

3 years ago

5, 8, II, 14 ...<br>The pattern is<br>Next 2 terms =

QveST [7]

Answer:

the next two terms are 17, 20

Step-by-step explanation:

add 3 to each term

Example

5+3 = 8

8+3= 11

11+3= 14 etc.

4 0

3 years ago

Read 2 more answers

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

This is my question!

tatuchka [14]

Answer:

Step-by-step explanation:

mode = 56

mean = 37.4

median = 35

6 0

3 years ago