3 years ago

Find the domain of f(x)= sqrt2-9x

Mathematics

1 answer:

victus00 [196]3 years ago

5 0

f(x) = $\sqrt{2 - 9x}$

2 - 9x ≥ 0

- 9x ≥ - 2

$\frac{-9x}{-9} \geq \frac{-2}{-9}$

x ≤ $\frac{2}{9}$

Interval Notation: (-∞, $\frac{2}{9}$ ]

You might be interested in

When a number is decreased by 8.3%, the result is 50. What is the original number to the nearest tenth?

lesya692 [45]

I'm pretty sure the answer would be 2.

Sorry if it's wrong

5 0

2 years ago

Read 2 more answers

jefferson park sits on one square city block 300 feet on each side. Sidewalks join opposite corners. About how long is each diag

wolverine [178]

About 425 feet, a^2*b^2=c^2, so 90,000+90,000=180,000 and the square root of 180,000 is about 424.

5 0

2 years ago

What is the slope line through (-9,-6) and (3,-9)?

chubhunter [2.5K]

Answer:

-1/4

Step-by-step explanation:

We can find the slope of the line given two points

m = (y2-y1)/(x2-x1)

= (-9- -6)/(3- -9)

= (-9+6)/(3+9)

=-3/12

= -1/4

8 0

2 years ago

If f(x)=|x+2| is changed to g(x)=3f(x)+4, how is the graph of the function transformed?

frozen [14]

Answer:

download photomath it will help u with any type of math

3 0

2 years ago

Geometry: Write a formal proof, ASAP!!!

yulyashka [42]

${\qquad\qquad\huge\underline{{\sf Answer}}}$

Here we go ~

As we can see in the figure, lines m and n are parallel to each other. so we can apply Angle pair properties ~

So,

$\qquad \sf \dashrightarrow \: \angle1 \cong \angle2$

[ they form corresponding angle pair ]

and it's given that :

$\qquad \sf \dashrightarrow \: \angle2 \cong\angle 3$

therefore, by using above information. we can conclude that :

$\qquad \sf \dashrightarrow \: \angle1 \cong \angle3$

7 0

1 year ago