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2 years ago

Convert the following into a FRACTION GREATER THAN 1. (Improper fractions) 3 3/7 & 3 5/6

Mathematics

1 answer:

Igoryamba2 years ago

7 0

Answer:

1. $\frac{24}{7}$
2. $\frac{21}{6}$

Step-by-step explanation:

Given the following questions:

$3\frac{3}{7} \\3\frac{5}{6}$

To convert mixed numbers into improper fractions we need to multiply the denominators by the whole numbers, and then add that answer to the numerator while of course keeping the denominator the same.

<u>Question one:</u>

$3\frac{3}{7} \\3\frac{3}{7} =7\times3=21+3=24=\frac{24}{7} \\=\frac{24}{7}$

<u>Question two:</u>

$3\frac{5}{6} \\3\frac{5}{6}=6\times3=18+3=21=\frac{21}{6} \\=\frac{21}{6}$

Your answers are "24/7, and 21/6."

Hope this helps.

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3 years ago

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4 years ago

Find the value(s) of k such that 4x² + kx + 4 = 0 has 1 rational solution

Blababa [14]

Answer:

If you mean only one rational solution, the answer is

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If you mean at least 1 rational solution, the answer is

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Step-by-step explanation:

$4x^2 + kx + 4 = 0$

Let's calculate the discriminant.

$\Delta = b^2 - 4ac$

$\Delta = k^2 -4 \cdot 4 \cdot 4$

$\Delta = k^2 -64$

Now, remember that:

$\text{If } \Delta > 0 : \text{2 Real solutions}$

$\text{If } \Delta = 0 : \text{1 Real solution}$

$\text{If } \Delta < 0 : \text{No Real solution}$

Therefore, I will just consider the first two cases.

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and

$k^2 -64 = 0$

$\boxed{\text{For } k^2 - 64 > 0}$

$k^2 > 64 \Longleftrightarrow k>\pm\sqrt{64} \Longleftrightarrow k\in (-\infty, -8)\cup(8, \infty)$

$\boxed{\text{For } k^2 - 64 = 0}$

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3 years ago

David is making his own strawberry yogurt. In his mixture, the number of strawberries is proportional to the amount of milk in c

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