All categories

3 years ago

Solve this equation.

Mathematics

2 answers:

Andrew [12]3 years ago

7 0

Answer:

Step-by-step explanation:

xz_007 [3.2K]3 years ago

4 0

$-\dfrac{2}{5}x-9 < \dfrac{9}{10}\qquad|\cdot10\\\\2(-2x)-90 < 9\qquad|+90\\\\-4x < 99\qquad|\text{change the signs}\\\\4x > -99\qquad|:4\\\\x > -\dfrac{99}{4}$

You might be interested in

Mrs.Vega brought a new aquarium for her turtles. How much space will the turtles have in the aquarium if the length is 5.2 ft, t

MrRissso [65]

Answer:

18.72 cubic feet

Step-by-step explanation:

Volume of the aquarium = lwh

l = 5.2 ft,
w = 1.8 ft
h = 2 ft

V = 5.2*1.8*2 = 18.72 cubic feet

3 0

3 years ago

Find the cost of tiling a rectangular plot of land 300m long and 150m wide at the rate of $6 per hundred square m

Ugo [173]

Answer:

hope this helps you.................

3 0

2 years ago

Cos (A+B) =1/5 COS (A-B) = 3/5 Then find tan (A-B)

Soloha48 [4]

Answer:

tan (A-B) = ± 4/3

Step-by-step explanation:

COS (A-B) = 3/5

COS² (A-B) = (3/5)² = 9/25 = 1 - sin² (A-B)

sin² (A-B) = 1 - 9/25 = 16/25

sin (A-B) = ± 4/5

tan (A-B) = sin (A-B) / cos (A-B) = (± 4/5) / (3/5) = ± 4/3

4 0

3 years ago

18-3 -18-3 -18-(-3) 18-(-3)

Marina CMI [18]

Answer:

1.) 15

2.) -21

3.) -15

4.) 21

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

a line whose perpendicular distance from the origin is 4 units and the slope of perpendicular is 2÷3. Find the equation of the l

GrogVix [38]

Answer:

$\huge\boxed{y=\dfrac{2}{3}x-\dfrac{4\sqrt{13}}{3}\ \vee\ y=\dfrac{2}{3}x+\dfrac{4\sqrt{13}}{3}}$

Step-by-step explanation:

The equation of a line:

$y=mx+b$

We have

$m=\dfrac{2}{3}$

substitute:

$y=\dfrac{2}{3}x+b$

The formula of a distance between a point and a line:

General form of a line:

$Ax+By+C=0$

Point:

$(x_0,\ y_0)$

Distance:

$d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+b^2}}$

Convert the equation:

$y=\dfrac{2}{3}x+b$ |subtract $y$ from both sides

$\dfrac{2}{3}x-y+b=0$ |multiply both sides by 3

$2x-3y+3b=0\to A=2,\ B=-3,\ C=3b$

Coordinates of the point:

$(0,\ 0)\to x_0=0,\ y_0=0$

substitute:

$d=4$

$4=\dfrac{|2\cdot0+(-3)\cdot0+3b|}{\sqrt{2^2+(-3)^2}}\\\\4=\dfrac{|3b|}{\sqrt{4+9}}$

$4=\dfrac{|3b|}{\sqrt{13}}\qquad|$ |multiply both sides by $\sqrt{13}$

$4\sqrt{13}=|3b|\iff3b=-4\sqrt{13}\ \vee\ 3b=4\sqrt{13}$ |divide both sides by 3

$b=-\dfrac{4\sqrt{13}}{3}\ \vee\ b=\dfrac{4\sqrt{13}}{3}$

Finally:

$y=\dfrac{2}{3}x-\dfrac{4\sqrt{13}}{3}\ \vee\ y=\dfrac{4\sqrt{13}}{3}$

4 0

3 years ago