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4 years ago

Melinda has ten fewer nickels than quarters.She has $5.20 in the bank.How many coins of each type does she have.

1 answer:

4 0

Well a quarter is = to 25 and a nickel is = to 5 so we know that there are fewer Nickels than quarters, therefore we know that we should you 4 quarters 5 times to build 5 dollars which is 20 quarters and use the nickels to build up 20 which would 4 since it be (5,10,15,20) counted by 5’s hope that hopes bby bops

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3 years ago

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Every morning, Halle goes to a school with a 1-liter bottle of water. She drinks 1/4 of the bottle before school starts and 2/3

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3 years ago

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

4 0

3 years ago