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djyliett [7]
2 years ago
10

State a counterexample to the conjecture below if a number is divisible by 6 and 2 then there’s is divisible by 12

Mathematics
2 answers:
Lisa [10]2 years ago
7 0
If there's is not divisible by 12 then the  number is not divisible by 6 and 2. Hope this helps.
siniylev [52]2 years ago
4 0
Any number that is divisible by 6 is already divisible by 2, but is not necessarily divisible by 12.

Counterexamples include: 6, 18, 30, 42, 54, and so on. You can find more by multiplying 6 by any odd number. However, multiplying 6 by an even number provides another "2" that would make it divisible by 12.
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A cook has 11 7/8 cups of flour. He uses 2/3 of his flour on Saturday morning. Estimate how many cups he used. Show how you esti
ziro4ka [17]

Answer:

11 (if you wanna know how many cups [which i mean by 2/3] it is 1)

Step-by-step explanation:

Well, to start, a cook has 11 7/8 cups of flour, right? Well, if you subtract 11 7/8 by 2/3 you get 11 5/24. As an estimate, you would get just plain 11 because there is three numbers when you estimate, 11, 11 1/2, or 12. 11 5/24 is more closer to 11 (because 1/2 would be 11 12/24, and 12 would be 11 13/24 or more until 12)

Hope it helped...

4 0
2 years ago
Read 2 more answers
What is a solution to the equation 3 / m + 3 - M / 3 - M equals m^2 + 9 / m^2-9?​
Mnenie [13.5K]

Answer: Last option.

Step-by-step explanation:

 Given the equation:

\frac{3}{m+3}-\frac{m}{3-m}=\frac{m^2+9}{m^2-9}

Follow these steps to solve it:

- Subtract the fractions on the left side of the equation:

\frac{3(3-m)-m(m+3)}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}\\\\\frac{9-3m-m^2-3m}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}\\\\\frac{-m^2-6m+9}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}

- Using the Difference of squares formula (a^2-b^2=(a+b)(a-b)) we can simplify the denominator of the right side of the equation:

\frac{-m^2-6m+9}{(m+3)(3-m)}=\frac{m^2+9}{(m+3)(m-3)}

- Multiply both sides of the equation by (m+3)(3-m) and simplify:

\frac{(-m^2-6m+9)(m+3)(3-m)}{(m+3)(3-m)}=\frac{(m^2+9)(m+3)(3-m)}{(m+3)(m-3)}\\\\-m^2-6m+9=\frac{(m^2+9)(3-m)}{(m-3)}

- Multiply both sides by m-3:

(-m^2-6m+9)(m-3)=\frac{(m^2+9)(3-m)(m-3)}{(m-3)}\\\\(-m^2-6m+9)(m-3)=(m^2+9)(3-m)

- Apply Distributive property and simplify:

(-m^2-6m+9)(m-3)=(m^2+9)(3-m)\\\\-m^3-6m^2+9m+3m^2+18m-27=3m^2+27-m^3-9m\\\\-m^3-3m^2+27m-27+m^3-3m^2+9m-27=0\\\\-6m^2+36m-54=0

- Divide both sides of the equation by -6:

\frac{-6m^2+36m-54}{-6}=\frac{0}{-6}\\\\m^2-6m+9=0

- Factor the equation and solve for "m":

(m-3)^2=0\\\\m=3

In order to verify it, you must substitute m=3 into the equation and solve it:

\frac{3}{3+3}-\frac{3}{3-3}=\frac{3^2+9}{3^2-9}\\\\\frac{3}{6}-\frac{3}{0}=\frac{18}{0}

<em>NO SOLUTION</em>

7 0
3 years ago
T is the midpoint of AE. AT=4x-3 and TE= 2x+7, find the length of AE
iris [78.8K]

If T is the midpoint, then that means its exactly 1/2 of the way between 2 points

this means we can set the 2 given values equal

4x - 3 = 2x + 7

first move like terms to the same side

2x = 10

x = 5

now that we know the length of x, we plug it back in to get the length

4x - 3 + 2x + 7 = 6x + 4

6(5) + 4 = 34

7 0
3 years ago
Find the product. √5(2+√7) A)2√5+√35 B)√45 C)2√5+2√7 D)√10+√35
OverLord2011 [107]

Answer:

\huge \boxed{2\sqrt{5} +\sqrt{35}}

Step-by-step explanation:

\sqrt{5} (2+\sqrt{7} )

Expand brackets.

\sqrt{5} (2)+\sqrt{5}(\sqrt{7} )

2\sqrt{5} +\sqrt{5} \sqrt{7}

Apply radical rule : \sqrt{a} \sqrt{b} =\sqrt{ab}

2\sqrt{5} +\sqrt{5 \times 7}

2\sqrt{5} +\sqrt{35}

7 0
3 years ago
Read 2 more answers
Find the value of b. Enter as an exact, simplified value:
Zanzabum

Answer: 3sqrt6

Step-by-step explanation:

6 0
2 years ago
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