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3 years ago

What are the solutions of the equation f(x) = g(x), given the following:

Mathematics

1 answer:

emmainna [20.7K]3 years ago

7 0

F(x) = g(x) for x = 1 and x = 3. Verify that for yourself.

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What does the constant term in the numerator represent?

Yakvenalex [24]

Answer:

A constant in math is a value that doesn't change. Instead, it's a fixed value. All numbers are considered constant terms.

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2 years ago

Eric has 4 pieces of clay he cut out each piece of clay into thirds how many 1/3 size pieces of clay does Eric have

Ivahew [28]

Answer:

ERIC HAD 12 PIECES OF CLAY LEFT

Step-by-step explanation:

1/3 + 4 = 12

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3 years ago

Round 2.71828 to the hearest thousandth.

Colt1911 [192]

Answer:

2.718

Step-by-step explanation:

The thousandths place is 3 to the right of the decimal (the 8).

Since the next number is 2, the number stays the same and does not round up.

Hope this helps :)

4 0

3 years ago

The perimeter of a stop sign is 99.2 inches. Each side of the stop sign is equal.

san4es73 [151]

Answer: The equation can be written as 8n = 99.2.

Step-by-step explanation:

In my answer, I used the variable n to distinguish itself as one side of the stop sign. 99.2 is the final perimeter in inches.

A stop sign is an octagonal shape (otherwise, it is made up of 8 equal sides). Therefore, with this information, you can interpret that the length of one side would be equal to the length of all of the other sides.

Hence, the answer can be represented as 8n = 99.2.

6 0

3 years ago

Which expression is equivalent to *picture attached*

DiKsa [7]

Answer:

The correct option is;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$

Step-by-step explanation:

The given expression is presented as follows;

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right )$

Which can be expanded into the following form;

$\sum\limits _{n = 1}^{50} \left (4\cdot n^2 + 3 \cdot n\right ) = 4 \times \sum\limits _{n = 1}^{50} \left n^2 + 3 \times\sum\limits _{n = 1}^{50} n$

From which we have;

$\sum\limits _{k = 1}^{n} \left k^2 = \dfrac{n \times (n+1) \times(2n+1)}{6}$

$\sum\limits _{k = 1}^{n} \left k = \dfrac{n \times (n+1) }{2}$

Therefore, substituting the value of n = 50 we have;

$\sum\limits _{n = 1}^{50} \left k^2 = \dfrac{50 \times (50+1) \times(2\cdot 50+1)}{6}$

$\sum\limits _{k = 1}^{50} \left k = \dfrac{50 \times (50+1) }{2}$

Which gives;

$4 \times \sum\limits _{n = 1}^{50} \left n^2 = 4 \times \dfrac{n \times (n+1) \times(2n+1)}{6} = 4 \times \dfrac{50 \times (50+1) \times(2 \times 50+1)}{6}$

$3 \times\sum\limits _{n = 1}^{50} n = 3 \times \dfrac{n \times (n+1) }{2} = 3 \times \dfrac{50 \times (51) }{2}$

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right ) = 4 \times \dfrac{50 \times (50+1) \times(2\times 50+1)}{6} +3 \times \dfrac{50 \times (51) }{2}$

Therefore, we have;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$ .

4 0

3 years ago