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3 years ago

Suppose two fractions are both less than 1. Can their sum be greater than 1? Greater than 2?

2 answers:

8 0

Hi there!
Yes, if the two fractions are less than one, their sum can be greater than one. Take the example of 3/4 and 3/4. It is less than 1, but has a sum if 1 1/2.
However, their sum cannot be greater than 2. For that to happen, there would have to be a sum of at least 1 and 1, which are not fractions.

Hope this helps :)

Juliette [100K]3 years ago

4 0

Their sum can be greater than two. There is nothing that states that they couldn't be. All it states it that the two factions sums are less than 1. I hoped this helps

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The exponent indicates how many times the ......... is used as a factor

Dimas [21]

Base or base number
for example 2^3 = 2 * 2 * 2 or x^2 = x * x

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3 years ago

Let f(x) = (x − 3)−2. Find all values of c in (1, 7) such that f(7) − f(1) = f '(c)(7 − 1). (Enter your answers as a comma-separ

Sidana [21]

Answer:

This contradicts the Mean Value Theorem since there exists a c on (1, 7) such that f '(c) = f(7) − f(1) (7 − 1) , but f is not continuous at x = 3

Step-by-step explanation:

The given function is

$f(x)=(x-3)^{-2}$

When we differentiate this function with respect to x, we get;

$f'(x)=-\frac{2}{(x-3)^3}$

We want to find all values of c in (1,7) such that f(7) − f(1) = f '(c)(7 − 1)

This implies that;

$0.06-0.25=-\frac{2}{(c-3)^3} (6)$

$-0.19=-\frac{12}{(c-3)^3}$

$(c-3)^3=\frac{-12}{-0.19}$

$(c-3)^3=63.15789$

$c-3=\sqrt[3]{63.15789}$

$c=3+\sqrt[3]{63.15789}$

$c=6.98$

If this function satisfies the Mean Value Theorem, then f must be continuous on [1,7] and differentiable on (1,7).

But f is not continuous at x=3, hence this hypothesis of the Mean Value Theorem is contradicted.

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3 years ago

Chase works as a welder making $21.50 per hour. He works 40 hours.

m_a_m_a [10]

What is his net weekly pay?

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2 years ago

Read 2 more answers

Write each expression as a single power of 10

swat32

Answer:

It is

${10}^{5}$

Step-by-step explanation:

$\frac{ {10}^{4}. {10}^{5} . {10}^{6} }{ {10}^{3} . {10}^{7} } = \frac{ {10}^{(4 + 5 + 6)} }{ {10}^{(3 + 7)} } \\ \\ = \frac{ {10}^{15} }{ {10}^{10} } \\ \\ = {10}^{(15 - 10)} \\ = {10}^{5}$

3 0

3 years ago

If Mike has m pencils and Jen has 3 more pencils than Mike, how many pencils does jen have?

Lera25 [3.4K]

Answer:

M + 3

Step-by-step explanation:

If Jen has 3 more pencils than mike, then you add the amount of pencils to however many mike has which leads to M+3

6 0

2 years ago