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Arlecino [84]
3 years ago
5

How many dogs do I have if I give two away

Mathematics
1 answer:
Helen [10]3 years ago
3 0

Answer:

Two less dogs

Your situation can be defined as x-2, where x was the number of dogs. Since the number of original dogs was not specified, two less dogs is the corrext answer

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Please can someone help me please
motikmotik

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Variable with constant, espression and exponent.

Step-by-step explanation:

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List all the numbers from 1 to 100 that have 13 as a factor
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13,26,39,52,65,78,91
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-X2 - x + 42
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Step-by-step explanation:

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3 years ago
Need step-by-step. Please answer correctly. This determines what grade I get. I will mark brainliest to the first person who ans
tatyana61 [14]

Answer:

  3√3

Step-by-step explanation:

For the problem shown here, your answer 3√3 is correct.

When there is a radical by itself in the denominator, you multiply numerator and denominator by a radical that results in the product being rational. For a square root, that will usually be the same square root:

  \dfrac{9}{\sqrt{3}}=\dfrac{9}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{9\sqrt{3}}{3}=\boxed{3\sqrt{3}}

__

If the problem has a sum in the denominator involving a square root, then you multiply numerator and denominator by the conjugate of that sum (the sum with the sign changed). This uses the special product "difference of squares" to eliminate the radical term.

<u>Example</u>:

  \dfrac{9}{2-\sqrt{3}}=\dfrac{9}{2-\sqrt{3}}\cdot\dfrac{2+\sqrt{3}}{2+\sqrt{3}}=\dfrac{9(2+\sqrt{3})}{2^2-(\sqrt{3})^2}=\dfrac{18+9\sqrt{3}}{4-3}\\\\=\boxed{18+9\sqrt{3}}

__

It is easy to demonstrate that none of the offered choices for this problem has the same value as 9/√3.

9/√3 ≈ 5.196. Offered choices have values of about 4.798, 1.732, 6.681, 23.196 -- none even close.

Please discuss this question with your teacher.

3 0
2 years ago
Sue drank 2/5 of her water bottle and sally drank 4/10 of her water bottle how much water did both girls drank?
stealth61 [152]

well, clearly the LCD from the denominators of 5 and 10 is just 10, thus

\bf \stackrel{\textit{Sue}}{\cfrac{2}{5}}+\stackrel{\textit{Sally}}{\cfrac{4}{10}}\implies \stackrel{\textit{LCD of 10}}{\cfrac{(2)2~~+~~(1)4}{10}}\implies \cfrac{2+4}{10}\implies \cfrac{6}{10}\implies \stackrel{\textit{simplified}}{\cfrac{3}{5}}

3 0
3 years ago
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