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3 years ago

A coordinate plane with a line passing through (negative 4, negative 4), (0, 0) and (4, 4).

Mathematics

2 answers:

antiseptic1488 [7]3 years ago

8 0

Answer:B

Step-by-step explanation: x - 1/2y = 0 just seems like the most reasonable answer.

Kisachek [45]3 years ago

5 0

Answer:

-x + y = 0 or x - y = 0

Step-by-step explanation:

The line passes through points (0, 0) and (4,4), then its slope is:

m = (4 - 0)/(4 - 0) = 1

Given that the line passes through (0, 0), its y-intercept is

b = 0

Using slope and y-intercept form:

y = mx + b

y = x

The standard form of a line is:

Ax + By = C

So, we have to put all variables on the left side, as follows:

-x + y = 0

Multiplying by (-1) at both sides:

x - y = 0

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The instructor thinks that the students may have answered the questions by guessing, so that the probability that any given answ

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Answer:

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3 years ago

Find the inverse of the function. f(x) = the cube root of quantity x divided by seven. - 9

elena-14-01-66 [18.8K]

To solve

replace f(x) with y
switch x and y
solve for y
replace y with f⁻¹(x)

so
I'm not sure if it is
$f(x)=\sqrt[3]{\frac{x}{7}}-9$ or
$f(x)=\sqrt[3]{\frac{x}{7}-9}$

first one
$f(x)=\sqrt[3]{\frac{x}{7}}-9$
replace f(x) with y
$y=\sqrt[3]{\frac{x}{7}}-9$
switch x and y
$x=\sqrt[3]{\frac{y}{7}}-9$
solve for y
$x+9=\sqrt[3]{\frac{y}{7}}$
$(x+9)^3=\frac{y}{7}$
$7(x+9)^3=y$
replace y with f⁻¹(x)
$f^{-1}(x)=7(x+9)^3$

2nd one
$f(x)=\sqrt[3]{\frac{x}{7}-9}$
replce f(x) with y
$y=\sqrt[3]{\frac{x}{7}-9}$
switch x and y
$x=\sqrt[3]{\frac{y}{7}-9}$
solve for y
$x^3=\frac{y}{7}-9$
$x^3+9=\frac{y}{7}$
$7(x^3+9)=y$
replace y with f⁻¹(x)
$f^{-1}(x)=7(x^3+9)$

if you meant $f(x)=\sqrt[3]{\frac{x}{7}}-9$ then the inverse is $f^{-1}(x)=7(x+9)^3$

if you meant $f(x)=\sqrt[3]{\frac{x}{7}-9}$ then the inverse is $f^{-1}(x)=7(x^3+9)$

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Answer:

97.75 seconds