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Alexeev081 [22]

2 years ago

12

Alex and glen share a 30 ounce bucket of clay by the end of the week alex has used 1/10 of the bucket and glen has used 2/5 of t

he bucket of clay how many ounces are left in the bucket

1 answer:

Mamont248 [21]2 years ago

6 0

Answer:

21 ounces

Step-by-step explanation:

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Answer:

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5x+9y+90

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Let f(x)=-5x-4 and g(x)=6x-7. find f(x) + g(x)

DIA [1.3K]

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10 months ago

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3 years ago

What is the domain (in interval notation) of the following functions?

alexgriva [62]

$1.\\g(x)=\frac{3}{5x-4}\\\\D:5x-4\neq0\to5x\neq4\ \ \ /:5\to x\neq\frac{4}{5}\to x\in\mathbb{R}\ \backslash\ \{\frac{4}{5}\}\\\\2.\\h(x)=\frac{\sqrt{x}}{x-5}\\\\D:x\geq0\ \wedge\ x-5\neq0\to x\geq0\ \wedge\ x\neq5\to x\in\left$

$3.\\f(x)=\frac{\sqrt{x}}{x^2-5x}\\\\D:x\geq0\ \wedge\ x^2-5x\neq0\to x\geq0\ \wedge\ x(x-5)\neq0\\\\\to x\geq0\ \wedge\ x\neq0\ \wedge\ x\neq5\to x\in\mathbb{R^+}\ \backslash\ \{5\}\\\\4.\\g(x)=\frac{\sqrt{x}+5}{x^2-x-20}\\\\D:x\geq0\ \wedge\ x^2-x-20\neq0\to x\geq0\ \wedge\ (x+4)(x-5)\neq0\\\\\to x\geq0\ \wedge\ x\neq-4\ \wedge\ x\neq5\to x\in\left$

$5.\\h(x)=\frac{3}{x^2+1}\\\\D:x^2+1\neq0\to x^2\neq-1\to x\in\mathbb{R}\\\\6.\\f(x)=\frac{\sqrt{x-2}}{x+1}\\\\D:x-2\geq0\ \wedge\ x+1\neq0\to x\geq2\ \wedge\ x\neq-1\to x\in\left$

$7.\\g(x)=\frac{x^2}{3x^2-x-2}\\\\D:3x^2-x-2\neq0\to (3x+2)(x-1)\neq0\to x\neq-\frac{2}{3}\ \wedge\ x\neq1\\\\\to x\in\mathbb{R}\ \backslash\ \{-\frac{2}{3};\ 1\}\\\\8.\\h(x)=3(x-4)^2-7\\\\D:x\in\mathbb{R}$

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2 years ago