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3 years ago

What is sum of all palindromic numbers from 1 to 100

Mathematics

2 answers:

Gekata [30.6K]3 years ago

7 0

Answer:

540

Step-by-step explanation:

0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99

Ber [7]3 years ago

5 0

Answer:

540

Step-by-step explanation:

Hey there!

Well we need to first find all the palindromic numbers,

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99

Add

= 540

Hope this helps :)

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3 years ago

Read 2 more answers

On #10 i need help to solve this. please and thank you.

valina [46]

10a. $f(x) = x - 4$
 $h(x) = \sqrt{x - 5}$
 $(f - h)(6) = (6 - 4) - (\sqrt{6 - 5})$
 $(f - h)(6) = 2 - \sqrt{1}$
 $(f - h)(6) = 2 - 1$
 $(f - h)(6) = 1$

10b. $f(x) = x - 4$
 $g(x) = \frac{1}{x - 3}$
 $(f * g)(x) = (x - 4)(\frac{1}{x - 3})$
 $(f * g)(x) = \frac{x - 4}{x - 3}$
 Domain: (-∞, 3) ∨ (3, ∞) {x|x ≠ 3}
 Interval Notation: (3 < x < 3), (3 > x > 3)

10c. $g(x) = \frac{1}{x - 3}$
 $h(x) = \sqrt{x - 5}$
 $(g * h)(x) = (\frac{1}{x - 3})(\sqrt{x - 5})$
 $(g * h)(x) = \frac{\sqrt{x - 5}}{x - 3}$

10d. $f(x) = x - 4$
 $g(x) = \frac{1}{x - 3}$
 $h(x) = \sqrt[3]{x - 7}$
 $h(x) = (f * g)(x)$
 $\sqrt[3]{x - 7} = (x - 4)(\frac{1}{x - 3})$
 $\sqrt[3]{x - 7} = \frac{x - 4}{x - 3}$
 $(\sqrt[3]{x - 7})^{3} = (\frac{x - 4}{x - 3})^{3}$
 $x - 7 = \frac{(x - 4)^{3}}{(x - 3)^{3}}$
 $(x - 3)^{3}(x - 7) = (x - 4)^{3}$
 $(x^{3} - 9x^2 + 27x + 27)(x - 7) = (x^{3} - 12x^{2} + 48x - 64)$
 $x^{4} - 16x^{3} + 90x^{2} - 216x + 189 = x^{3} - 12x^{2} + 48x - 64$
 $x^{4} + 90x^{2} - 216x + 189 = 17x^{3} - 12x^{2} + 48x - 64$
 $x^{4} + 102x^{2} - 216x + 189 = 17x^{3} + 48x - 64$
 $x^{4} + 102x^{2} + 189 = 17x^{3} + 264x - 64$
 $x^{4} + 102x^{2} + 253 = 17x^{3} + 264x$
 $x^{4} - 17x^{3} + 102x^{2} - 264x + 253 = 0$
 $x = 4\ or\ 8$