Given: F(x) = 2x2+ 1, G(x) = 2x - 1, H(x) = x F(3) + G(4) - 2H(5) = 16 26 34

Mathematics

1 answer:

Roman55 [17]3 years ago

3 0

Answer:

Step-by-step explanation:

Put the numbers where the variables are and evaluate:

F(3) = 2(3)^2 +1 = 19

G(4) = 2(4) -1 = 7

H(5) = 5

Then the desired sum is ...

F(3) +G(4) -2H(5) = 19 +7 -2(5) = 16

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Answer:

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Step-by-step explanation:

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Need help with this question plZ

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Without using a calculator, order the following expressions from least to greatest.

LekaFEV [45]

Answer:

$\displaystyle \large{\frac{11}{3} < \sqrt{17} < \sqrt{20}}$

Step-by-step explanation:

( 1 ) 11/3

$\displaystyle \large{\frac{11}{3}}$ can be converted to mixed fractions to $\displaystyle \large{3\frac{2}{3}}$

Therefore, 11/3 is at least around 3, almost 4. It could be more than 3.5 by approximation but not exactly 4.

( 2 ) √20

$\displaystyle \large{\sqrt{20}=\sqrt{5\cdot 4}}\\\displaystyle \large{\sqrt{20}=\sqrt{5\cdot 2\cdot 2}}\\\displaystyle \large{\sqrt{20}=2\sqrt{5}}$

We know that $\displaystyle \large{\sqrt{4}=2}$ so $\displaystyle \large{\sqrt{5} > \sqrt{4}}$ which means approximately, $\displaystyle \large{\sqrt{20}}$ can be either at least 4.3 up to 4.4

( 3 ) √17

We know that $\displaystyle \large{\sqrt{16} = 4}$ so $\displaystyle \large{\sqrt{17}}$ can be at least 4.1 up to 4.2, but it’s obvious that the square root of 17 is less than square root of 20 since $\displaystyle \large{\sqrt{a} > \sqrt{b}}$ for a > b