3 years ago

Graph the inequality 8x>3y

Mathematics

1 answer:

Alex Ar [27]3 years ago

4 0

I added a picture of what the graph should look like, the coordinates are (2,5)<span />

You might be interested in

484,168 in expanded form

liubo4ka [24]

400,000+80,000+4,000+100+60+8

8 0

3 years ago

Read 2 more answers

Maria solved the equation -5 n = -12. Her answer was n = 2. She is fairly confident about her answer, but wants to do a quick ch

lyudmila [28]

It would be the fourth answer -6(2)=-13 I think

4 0

3 years ago

Please help with the problem below! Thank you

Gwar [14]

Answer:

$d \le -3$

Step-by-step explanation:

The given inequality is;

$\frac{d-3}{10} \ge \frac{2d+3}{5}+\frac{d+3}{3}$

Clear the fraction by multiply through by 30;

$3(d-3) \ge 6(2d+3)+10(d+3)$

Expand;

$3d-9\ge 12d+18+10d+30$

Group like terms;

$3d-12d-10d \ge 18+30+9$

$-19d \ge 57$

Divide through by -19 and reverse the sign.

$d \le -3$

6 0

4 years ago

Read 2 more answers

Divide -2x3-5x2+4x+2 by x+2

lilavasa [31]

It would be -26

Its very easy But you will get better =)

4 0

3 years ago

If f(x)=ln x and g(x)=e^x+1 find f(g(2))-g(f(e)) please help!!

Readme [11.4K]

Answer:

Step-by-step explanation:

Given f(x) = ln and g(x) = e^x + 1 to get f(g(2))-g(f(e)), we need to first find the composite function f(g(x)) and g(f(x)).

For f(g(x));

f(g(x)) = f(e^x + 1)

substitute x for e^x + 1 in f(x)

f(g(x)) = ln (e^x + 1)

f(g(2)) = ln (e^2 + 1)

For g(f(x));

g(f(x)) = g(ln x)

substitute x for ln x in g(x)

g(f(x)) = e^lnx + 1

g(f(x)) = x+1

g(f(e)) = e+1

f(g(2))-g(f(e)) = ln (e^2 + 1) - (e+ 1)

5 0

3 years ago