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mina [271]
4 years ago
6

1/x<4x

; 4x" alt=" \frac{1}{x} < 4x" align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
liraira [26]4 years ago
6 0

Answer:

x>\frac{1}{2} \\x

Step-by-step explanation:

If we solve for 'x' variable, its recommended to multiply 'x' variable in both sides of inequality:

we have \frac{1}{4}< x^{2} \\

This is a quadratic equation, two answer are going to be obtained from here.

since \sqrt{x^{2} } = |x|

Applying square roots to both sides of inequality sign we wil have the following

\sqrt{\frac{1}{4} } < \sqrt{x^{2} }

This leaves to the following

\frac{1}{4}

remember that |x|= +/ - x

so

x>\frac{1}{2} \\x

You might be interested in
Form the union for the following sets.
Readme [11.4K]
The union of the sets are the values that appear in both sets when plotted in a Venn diagram and if I'm not mistaking, I think the right answer is (B)
3 0
3 years ago
Read 2 more answers
If angle 4 is 52.7, what are the measures of angles 1,2, and 3? ​
shusha [124]

Answer:

  1. 127.3
  2. 52.7
  3. 127.3
  4. 52.7

Step-by-step explanation:

Since we know the angle measure of angle 4, we already know that angle 2 will have the same measure according to do the vertical angle theorem. Now to find angles 1 and 3, we can make an equation and solve for x (supplementary angles).

52.7 + x = 180

x = 127.3

Best of Luck!

6 0
3 years ago
find the surface area of a right regular hexagonal pyramid with sides 3 cm and slant heights 6cm. show all of your work.​
DedPeter [7]

Answer:

The surface area of right regular hexagonal pyramid = 82.222 cm³

Step-by-step explanation:

Given as , for regular hexagonal pyramid :

The of base side = 3 cm

The slant heights = 6 cm

Now ,

The surface area of right regular hexagonal pyramid = \frac{3\sqrt{3}}{2}\times a^{2} + 3 a \sqrt{h^{2}+ 3\times \frac{a^{2}}{4}}

Where a is the base side

And h is the slant height

So, The surface area of right regular hexagonal pyramid = \frac{3\sqrt{3}}{2}\times 3^{2} + 3 \times 3 \sqrt{6^{2}+ 3\times \frac{3^{2}}{4}}

Or, The surface area of right regular hexagonal pyramid = \frac{3\sqrt{3}}{2}\times 9 + 9 \sqrt{36+ 3\times \frac{9}{4}}

Or,  The surface area of right regular hexagonal pyramid = 23.38 + 9 × \frac{171}{4}

∴  The surface area of right regular hexagonal pyramid = 23.38 + 9 × 6.538

I.e The surface area of right regular hexagonal pyramid = 23.38 + 58.842

So,  The surface area of right regular hexagonal pyramid = 82.222 cm³ Answer

6 0
4 years ago
Change the subject of the equation v = u + at to a.
emmasim [6.3K]
<h3>Answer ↓</h3>

<h3>Calculations ↓</h3>

In order to make a the subject of this equation , we need to get a by itself .

The current equation is :

v = u + at

Subtract u on both sides :

v-u=at

Now, divide by t on both sides :

v-u/t=a

<h3>So the formula looks like ↓</h3>

\boxed{\\\begin{minipage}{3cm}Equation \\ a=$\displaystyle\frac{v-u}{t} $ \\ \end{minipage}}

hope helpful ~

8 0
3 years ago
H = vt - 16t^2, solve for v
Sophie [7]
H=vt-16t²
vt-16t²=H
vt=16t²+H
v=(16t²+H)/t
v=16(t²/t)+H/t
v=16t+H/t

Answer: the answer would be: v=16t+H/t
4 0
4 years ago
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