3 years ago

-3(2r-4) > 2(5-3r) Whats the inequality?

Mathematics

2 answers:

Ksenya-84 [330]3 years ago

7 0

-6r=12>10-6r
+6r +6r
12>10

VLD [36.1K]3 years ago

4 0

-3 (2r - 4) > 2 (5 - 3r)
-6r + 12 > 10 - 6r
12 > 10

There is no solution

You might be interested in

Solve for x<br> x = 5<br> x = 18<br> x= 30°<br> x = 36

Norma-Jean [14]

Answer:

x=36

Step-by-step explanation:

5x=180

x=180/5=36

8 0

3 years ago

I really need someone to answer this

DENIUS [597]

Answer:

809.69

Step-by-step explanation:

747.98(.0825)=61.708

747.98+61.708=809.688

3 0

2 years ago

A fair coin is tossed 5 times in a row. The exact probability of the coin landing heads exactly 2 times is?

OLga [1]

$|\Omega|=2^5=32\\|A|=10\\\\P(A)=\dfrac{10}{32}=\dfrac{5}{16}$

5 0

3 years ago

An equation of an ellipse is given. y2 = 1 − 3x2 (a) Find the vertices, foci, and eccentricity of the ellipse. vertex (x, y) = (

oksian1 [2.3K]

Answer:

Step-by-step explanation:

Given

$y^2=1-3x^2$

$3x^2+y^2=1$

$\frac{x^2}{(\frac{1}{\sqrt{3}})^2}+\frac{y^2}{1}=1$

therefore it is a vertical ellipse

thus a=1

$b=\frac{1}{\sqrt{3}}$

eccentricity of Ellipse

$e^2=1-\frac{b^2}{a^2}$

$e^2=1-\frac{1}{(\sqrt{3})^2}$

$e^2=1-\frac{1}{3}$

$e^2=\frac{2}{3}$

$e=\sqrt{\frac{2}{3}}$

Focii are $(0,ae)$ and $(0,-ae)$

$ae=1\times \sqrt{\frac{2}{3}}$

thus focii are $(0,\sqrt{\frac{2}{3}})$ & $(0,-\sqrt{\frac{2}{3}})$

(b) Length of major axis $=2a=2\times 1$

length of minor axis $=2b=2\times \sqrt{\frac{2}{3}}=2\cdot \sqrt{\frac{2}{3}}$

8 0

3 years ago

3.24 as a mixed number

UkoKoshka [18]

3 6/25 would be your answer!!!

8 0

3 years ago