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3 years ago

What is the equation of the line passing through points (1,1) and (3,-5)

Mathematics

1 answer:

seraphim [82]3 years ago

5 0

Answer:

m=-3

Step-by-step explanation:

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Need to know now!!! giving 10 points what -6+8

Simora [160]

Answer:

the answer is 2...................

Step-by-step explanation:

-6+8=2

7 0

3 years ago

Read 2 more answers

Find the perimeter of the figure 35 in. 34 in. 70 in. 84 in.

Yuri [45]

Answer:

p=8in+8in+12in+7in

=35 in

8 0

3 years ago

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Thanks guys for the help. This one and one more and I’m done!

Karolina [17]

Answer: $60n^{3}$

Step-by-step explanation:

To find the least common multiply, you must descompose 12 and 15 into their prime factors, as you can see below:

12=2*2*3=2²*3

15=3*5

Choose the common and non common numbers with their greastest exponents:

3*5*2²=60

Now you must choose the common and non common variables with their greastest exponents:

n³

Therefore, you can conclude that the least common multiply is:

$60n^{3}$

8 0

4 years ago

Read 2 more answers

Marissa bought four bottles of water Each bottle of water is 0.95 cents Write an equation with the same product as the total cos

zysi [14]

Answer: Total cost = Cost of each bottle × Number of bottles

Step-by-step explanation: If each bottle of water Marissa bought cost 0.95cents. This means four bottles will cost = 0.95 + 0.95 + 0.95 + 0.95 = 3.80 cents

This can be written simply as 0.95 × 4 = 3.80 cents

Therefore, a general equation for calculating the total cost of bottle water will be;

Total cost = Cost of each bottle × Number of bottles.

8 0

3 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