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2 years ago

Please help I will give you a brainliest

Mathematics

1 answer:

LUCKY_DIMON [66]2 years ago

4 0

Look at the picture attached

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Evaluate the expression

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$\frac{16^{6/4}}{16^{2/2} }$

$\frac{64}{16}$

$4$

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2 years ago

If 5/8 is multiplied by 4/5 will the product be greater than the two factors? Explain.

STALIN [3.7K]

Answer:

Step-by-step explanation:

4/5 is less than 1, so multiplying 5/8 by 4/5 results in a product smaller than the two factors.

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2 years ago

Math linear functions

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Answer:

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3 years ago

How to solve - 10+5x=7x-4

Triss [41]

10+5x=7x-4
1. simplify 7x - 5x
10=2x-4

2. add four to both sides
10+4=2x
+ 4 4
14=2x

3. divide 14 to both sides
14=2x

x=7

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3 years ago

Prove 2√(x) + 1/√(x + 1) <= 2√(x+1) for all x in [0,inf)

EleoNora [17]

Start by multiplying each side of the inequality by $\sqrt{x + 1}$ and simplifying:

$2 \sqrt x + \frac{1}{\sqrt{x+1}} \leq 2 \sqrt{x + 1}$
$(2 \sqrt x + \frac{1}{\sqrt{x+1}})(\sqrt{x + 1}) \leq (2 \sqrt{x + 1})(\sqrt{x + 1})$
$2 \sqrt{x(x + 1)} + 1 \leq 2(x + 1)$
$2 \sqrt{x^2 + x} + 1 \leq 2x + 2$
$2 \sqrt{x^2 + x} \leq 2x + 1$
$\sqrt{x^2 + x} \leq x + \frac{1}{2}$

From here, we can square both sides to get

$x^2 + x \leq (x + \frac{1}{2})^2$
$x^2 + x \leq x^2 + x + \frac{1}{4}$
$0 \leq \frac{1}{4}$ , which is always true.

3 0

2 years ago