All categories

3 years ago

Solve 2x - 8 + 3x > 5x - 4 also if you can plz explain

Mathematics

1 answer:

WITCHER [35]3 years ago

7 0

Answer:

No solution

Step-by-step explanation:

2x - 8 + 3x > 5x - 4

2x + 3x - 8 > 5x - 4

5x - 8 > 5x - 4

-5x +8 > -5x +8

----------------------

0x > 4

This is no solution, as there is no 'x' value.

(0x just means there isn't anything left, I put that in order to show there wasn't anything.)

You might be interested in

Does this graph show a function?

Montano1993 [528]

9514 1404 393

Answer:

Step-by-step explanation:

A vertical line intersects the graph in <em>more than one place</em>. As a consequence, we say the graph "fails the vertical line test." When the test fails, the graph is not a graph of a function.

8 0

2 years ago

Solve irrational equation pls

rusak2 [61]

$\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)$

$
\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\$
$
4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
(x-1)^2(3x^2-28)=0\\
x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\$

There's one more condition I forgot about
$-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\$

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}$

3 0

2 years ago

Can somebody help me. Will mark brainliest

MariettaO [177]

Answer:

the first two reason are given then the last one is cpctc

5 0

2 years ago

Solve for me please this is trigonometry

Vladimir79 [104]

Answer:

$x= 23.4$

Step-by-step explanation:

In a right triangle the tangent of an angle is the ratio of the opposite length over the adjacent*. In formulas, $tan 54° = x/17$ Solving for x, using a calculator for the tangent, $x= 23.4$

*I promise I didn't just spellcheck that.

3 0

1 year ago

A cone-shaped paper drinking cup is to be made to hold 33 cm3 of water. find the height and radius of the cup that will use the

alekssr [168]

The height and radius that will give us the smallest amount of paper, should have should have the perfect dimension, in that the diameter should be equal to the height. thus let the diameter be x, the height will be x and the radius will be x/2
thus the volume of the cone will be:
V=1/3πr^2h
=1/3*π*(x/2)^2*x
=0.262x^3
hence the value of x will be:
33=0.262x^3
x^3=125.95
x=(125.95)^(1/3)
x=5.0126=5.01 cm
thus the diameter=height=5.01cm

6 0

3 years ago